\(\text { If } A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right], \text { then adj } A=\left[\begin{array}{rr} d & -b \\ -c & a \end{array}\right] \text { . }\)
So, Adj \(A=\begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix}\)

\(\text { If } a \text { and } \vec{b} \text { are perpendicular, then } \vec{a} \cdot \vec{b}=0\)
\(\Rightarrow 2-2 \lambda+3=0 \Rightarrow \lambda=\frac{5}{2}\)

Direction cosines are \({\pm {1\over \sqrt3}},{\pm {1\over \sqrt3}},{\pm {1\over \sqrt3}}\)

\(A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \ B=\left[ \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right] \)

= 5

Given, P(A) = 0.3 and P(B) = 0.6
Now, \(\begin{aligned} P\left(A^{\prime} \cap B^{\prime}\right) & =P(A \cup B)^{\prime} \end{aligned}\)
\(\begin{aligned} =1-P[A \cup B] \end{aligned}\)
\(\begin{aligned} =1-[P(A)+P(B)-P(A \cap B)] \end{aligned}\)
\(\begin{aligned} =1-\{0.3+0.6-0.3 \times 0.6\} \end{aligned}\)
[\(\because\)  A and B are independent events \(\therefore P(A \cap B)=P(A) P(B)]\)
= 1- {0.9 - 0.18}
= 1 - {0.72} = 0.28

Let A_T : Event that A speaks truth
and B_T : Event that B speaks truth.
Given, \(\begin{aligned} P\left(A_T\right)=\frac{80}{100}=\frac{4}{5} \end{aligned}\)
\(\begin{aligned} P\left(B_T\right)=\frac{90}{100}=\frac{9}{10} \end{aligned}\)
P(agree) = P(both speaking truth or both telling lie)
\(\begin{aligned} =P\left(A_T B_T \text { or } \bar{A}_T \bar{B}_T\right) \end{aligned}\)
\(\begin{aligned} =P\left(A_T\right) P\left(B_T\right) \text { or } P\left(\bar{A}_T\right) P\left(\bar{B}_T\right) \end{aligned}\)
\(\begin{aligned} =\left(\frac{4}{5}\right)\left(\frac{9}{10}\right)+\left(\frac{1}{5}\right)\left(\frac{1}{10}\right) \end{aligned}\)
\(\begin{aligned} =\frac{36+1}{50}=\frac{37}{50}=\frac{74}{100}=74 \% \end{aligned}\)

-15

\(M_{11}=\begin{vmatrix} 0&4\\5&-7\end{vmatrix}=-0-20=-20\)
\(A_{11}=(-1)^{1+1}M_{11}=(-1)^2(-20)=-20\)
\(M_{12}=\begin{vmatrix}6&4\\1&-7 \end{vmatrix}=-42-4=-46\)
\(A_{12}=(-1)^{1+2}M_{12}=(-1)^3(-46)=(-1)(-46)=46\)
\(M_{13}=\begin{vmatrix}6&0\\1&5 \end{vmatrix}=30-0=30\)
\(A_{13}=(-1)^{1+3}M_{13}=(-1)^4(30)=30\)
\(M_{21}=\begin{vmatrix} -3&5\\5&-7\end{vmatrix}=21-25=-4\)
\(A_{21}=(-1)^{ 2+1}M_{21}=(-1)^3(-4=(-1 )(-4)=4)\)
\(M_{22}=\begin{vmatrix} 2&5\\1&-7\end{vmatrix}=-14-15=-19\)
\(A_{22}=(-1)^{2+2}M_{22}=(1)^4(-19)=-19\)
\(M_{23}=\begin{vmatrix} 2&-3\\1&5\end{vmatrix}=10+3=13\)
\(A_{23}=(-1)^{2+3}M_{23}=(-1)^513=-13\)
\(M_{31}=\begin{vmatrix}-3&5\\0&4 \end{vmatrix}=-12-0=-12\)
\(A_{31}=(-1)^{3+1}M_{31}=(-1)^4(-12)=-12\)
\(M_{32}=\begin{vmatrix} 2&5\\6&4\end{vmatrix}=8-30=-22\)
\(A_{32}=(-1)^{3+2}M_{32}=(-1)^5(-22)=(-1)(-22)=22\)
\(M_{33}=\begin{vmatrix} 2&-3\\6&0\end{vmatrix}=0+18=18\)
\(A_{33}=(-1)^{3+3}M_{33}=(-1)^6(18)=18\)
(ii)\(a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{32}\)
\(=(2)(-12)+(-3)(22)+(5)(18)=-24-66+90=0\)

(i) \(P(E/F)=\frac { P(E\cap F) }{ P(F) } =\frac { 0.2 }{ 0.3 } =\frac { 2 }{ 3 } \)
(ii) \(P(F/E)=\frac { P(E\cap F) }{ P(E) } =\frac { 0.2 }{ 0.6 } =\frac { 1 }{ 3 }\)

\(\text { Here } \vec{b}_{1}=\hat{i}+2 \hat{j}+2 \hat{k} \text { and } \vec{b}_{2}=3 \hat{i}+2 \hat{j}+6 \hat{k}\)
\(\text { The angle } \theta \text { between the two lines is given by }\)
\(\cos \theta =\left|\frac{\vec{b}_{1} \cdot \vec{b}_{2}}{\left|\vec{b}_{1}\right|\left|\vec{b}_{2}\right|}\right|=\left|\frac{(\hat{i}+2 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}+2 \hat{j}+6 \hat{k})}{\sqrt{1+4+4} \sqrt{9+4+36}}\right| \\ \)
\(=\left|\frac{3+4+12}{3 \times 7}\right|=\frac{19}{21} \)
\(\text { Hence }\theta=\cos ^{-1}\left(\frac{19}{21}\right) \)

Given equation of lines are
\(r=(4 \hat{i}-\hat{j})+\lambda(\hat{i}+2 \hat{j}-3 \hat{k})\)  ..(i) 
and \(r=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+4 \hat{j}-5 \hat{k})\) ...(ii)
On comparing Eqs. (i) and (ii) with \(r=a_{1}+\lambda b_{1}\) and \(r=a_{2}+\mu b_{2}\) respectively, we get
and \(a_{2}=\hat{i}-\hat{j}+2 \hat{k}, b_{2}=2 \hat{i}+4 \hat{j}-5 \hat{k}\) 
Here \(a_{2}-a_{1}=-3 \hat{i}+2 \hat{k}\) 
and \(b_{1} \times b_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{array}\right|\) 
\(=\hat{i}(-10+12)-\hat{j}(-5+6)+\hat{k}(4-4)\) 
\(=2 \hat{i}-\hat{j}\) 
\(\Rightarrow \left|b_{1} \times b_{2}\right|=\sqrt{2^{2}+(-1)^{2}}=\sqrt{4+1}=\sqrt{5}\) 
Now, the shortest distance between the given lines is given by
\(d=\frac{\left|\left(b_{1} \times b_{2}\right) \cdot\left(a_{2}-a_{1}\right)\right|}{\left|b_{1} \times b_{2}\right|}=\frac{|(2 \hat{i}-\hat{j}) \cdot(-3 \hat{i}+2 \hat{k})|}{\sqrt{5}}\) 
\(=\frac{|-6|}{\sqrt{5}}=\frac{6}{\sqrt{5}} \text { units }\)

\(P(A \cup B)=P(A)+P(B)-P(A \cap B)\) 
\(\Rightarrow \frac{7}{11}=\frac{6}{11}+\frac{5}{11}-P(A \cap B)\) 
\(\Rightarrow P(A \cap B)=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}\) 
\(\therefore P(A \cap B)=\frac{4}{11}\)

Using equality of matrices, we have
\(x+2=0, \quad 2 y+x=-7, \quad z-4=3 \text { and } 4 w-6=w \)
\(\Rightarrow x=-2, \quad 2 y-2=-7 \)
\(\Rightarrow y=-\frac{5}{2}, z=7 \text { and } w=2 \)

(i)Let 'x', 'y' and 'z' be the amount invested in three investments.
Then
x + y + z = 65000  ......(1)
\({6x\over100}+{8y\over100}+{9z\over100}=4800\)
\(\Rightarrow\) 6x + 8y + 9z = 480000  ...(2)
\({9z\over100}=600+{8y\over100}\)
\(\Rightarrow\) 0x - 8y + 9z = 60000
These can be written as AX = B where:
\(A=\begin{bmatrix} 1&1&1\\6&8&9\\0&-8&9\end{bmatrix},X=\begin{bmatrix} x\\y\\z\end{bmatrix}\)
and \(B=\begin{bmatrix} 650000\\480000\\60000\end{bmatrix}\)
(ii) Now \(|A|=\begin{bmatrix}1&1&1\\6&8&9\\0&-8&9 \end{bmatrix}\)
= 1.(72 + 72) - 6(9+8)
= 144 - 102 = 42 ≠ 0
Hence, the equations have a unique solution.
(iii) No.We are not fools because most of such companies are frauds.

Let E₁: Event selecting bag with 4 red & 4 black balls
E₂: Event selecting bag with 2 red & 6 black balls
A: Event selecting 2 red balls without replacement
Then, P(E₁) = P(E₂) = \(\frac { 1 }{ 2 } \)
\(P(A/{ E }_{ 1 })=\frac { 4_{ C_{ 2 } } }{ { 8 }_{ C_{ 2 } } } =\frac { 3 }{ 14 } ,\)
\(P(A/{ E }_{ 2 })=\frac { { 2 }_{ { C }_{ 2 } } }{ { 8 }_{ { C }_{ 2 } } } =\frac { 1 }{ 28 } .\)
\(P\left( \frac { { E }_{ 1 } }{ A } \right) =\frac { P({ E }_{ 1 }).P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 }).P(A/{ E }_{ 1 })+P({ E }_{ 2 }).P(A/{ E }_{ 2 }) } \)
\(=\frac { \frac { 1 }{ 2 } .\frac { 3 }{ 14 } }{ \frac { 1 }{ 2 } .\frac { 3 }{ 14 } +\frac { 1 }{ 2 } .\frac { 1 }{ 28 } } =\frac { 6 }{ 7 } \)

The lines which are neither intersecting nor parallel are called skew lines.
The given vector equations of two skew lines are
\(\vec{r}=8 \hat{i}-9 \hat{j}+10 \hat{k}+\lambda(3 \hat{i}-16 \hat{j}+7 \hat{k})\)       ...(i)
and \(\vec{r}=15 \hat{i}+29 \hat{j}+5 \hat{k}+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k})\) ...(ii)
Here, \(\vec{a}_1=8 \hat{i}-9 \hat{j}+10 \hat{k} ; \vec{a}_2=15 \hat{i}+29 \hat{j}+5 \hat{k}\)
and \(\overrightarrow{b_1}=3 \hat{i}-16 \hat{j}+7 \hat{k} ; \quad \overrightarrow{b_2}=3 \hat{i}+8 \hat{j}-5 \hat{k}\)
Now, \(\overrightarrow{a_2}-\overrightarrow{a_1}=(15-8) \hat{i}+(29+9) \hat{j}+(5-10) \hat{k}\)
\(\begin{aligned} &=7 \hat{i}+38 \hat{j}-5 \hat{k} \end{aligned}\)
\(\begin{aligned} \therefore \quad \vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{array}\right| \end{aligned}\)
\(\begin{aligned} =\hat{i}(80-56)-\hat{j}(-15-21)+\hat{k}(24+48) \end{aligned}\)
\(\begin{aligned} =24 \hat{i}+36 \hat{j}+72 \hat{k} \end{aligned}\)
\(\begin{aligned} \text { and }\left(\vec{b}_1\right. & \left.\times \vec{b}_2\right) \cdot\left(\overrightarrow{a_2}-\vec{a}_1\right) \end{aligned}\)
\(\begin{aligned} =(24 \hat{i}+36 \hat{j}+72 \hat{k}) \cdot(7 \hat{i}+38 \hat{j}-5 \hat{k}) \end{aligned}\)
= 168 + 1368 - 360 = 1176
\(\therefore\) Required shortest distance
\(\begin{aligned} d & =\left|\frac{\left(\overrightarrow{b_1} \times \vec{b}_2\right) \cdot\left(\overrightarrow{a_2}-\vec{a}_1\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right| \end{aligned}\)
\(\begin{aligned} =\left|\frac{1176}{\sqrt{24^2+36^2+72^2}}\right| \end{aligned}\)
\(\begin{aligned} =\frac{1176}{\sqrt{7056}} \end{aligned}\)
\(\begin{aligned} d & =\frac{1176}{84}=14 \text { units } \end{aligned}\)

Let \(B=\left[\begin{array}{ccc} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{array}\right] \text { and } A=\left[\begin{array}{crc} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{array}\right]\)
\(\therefore \quad B A=\left[\begin{array}{ccc} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{array}\right]\left[\begin{array}{ccc} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{array}\right]\)
\(=\left[\begin{array}{lll} -6+28+45 & 17+10-27 & 13-16+3 \\ -12+42-30 & 34+15+18 & 26-24-2 \\ -18-42+60 & 51-15-36 & 39+24+4 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 67 & 0 & 0 \\ 0 & 67 & 0 \\ 0 & 0 & 67 \end{array}\right]=67\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=67 I\)
\(\Rightarrow\) BA = 67 I
B^-1BA = B^-167     [Pre-multiplying both sides by B^-1]
\(\Rightarrow\) A = B^-167                                [\(\because\)  B^-1 = B = I]
\(\Rightarrow \quad B^{-1}=\frac{1}{67} A=\frac{1}{67}\left[\begin{array}{ccc} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{array}\right]\)
Given system of equations can be written in matrix form as
BX = C
\(\Rightarrow\)         X = B^-1C
where \(\beta=\left[\begin{array}{ccc} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{array}\right], C=\left[\begin{array}{c} -4 \\ 2 \\ 11 \end{array}\right] \text { and } X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\)
\(\therefore \quad X=\frac{1}{67}\left[\begin{array}{ccc} -6 & 17 & 13 \\ 14 & 5 & -8 \\ 15 & 9 & -1 \end{array}\right]\left[\begin{array}{c} -4 \\ 2 \\ 11 \end{array}\right]\)
\(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{67}\left[\begin{array}{l} 24+34+143 \\ -56+10-88 \\ +60+18-11 \end{array}\right]\)
\(=\frac{1}{67}\left[\begin{array}{c} 201 \\ -134 \\ +67 \end{array}\right]=\left[\begin{array}{c} 3 \\ -2 \\ 1 \end{array}\right]\)
On comparing corresponding elements, we get
x = 3, y = -2, z = 1

Given, equations of lines are
\(\begin{aligned} \vec{r}=(8 \hat{i}-19 \hat{j}+10 \hat{k})+\lambda(3 \hat{i}-16 \hat{j}+7 \hat{k}) \end{aligned}\)
and \(\begin{aligned} \vec{r}=(15 \hat{i}+29 \hat{j}+5 \hat{k})+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k}) \end{aligned}\)
On comparing with vector form of equation of a line,
i.e., \(\vec{r}=\vec{a}+\lambda \vec{b}\) we get
\(\vec{b}_1=3 \hat{i}-16 \hat{j}+7 \hat{k} \text { and } \vec{b}_2=3 \hat{i}+8 \hat{j}-5 \hat{k}\)
Now, we determine
\(\begin{aligned} \vec{b} & =\vec{b}_1 \times \vec{b}_2=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{array}\right| \end{aligned}\)
\(=\hat{i}(80-56)-\hat{j}(-15-21)+\hat{k}(24+48) \)
\(=24 \hat{i}+36 \hat{j}+72 \hat{k}=12(2 \hat{i}+3 \hat{j}+6 \hat{k})\)
Since, the required line is perpendicular to the given lines. So, it is parallel to \(\vec{b}_1 \times \vec{b}_2\). Now, the equation of a line passing through the point (1, 2, -4) and parallel to \(24 \hat{i}+36 \hat{j}+72 \hat{k} \text { or }(2 \hat{i}+3 \hat{j}+6 \hat{k})\) is
\(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\)
which is required vector equation of a line.
For cartesian equation, put \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\), we get
\(x \hat{i}+y \hat{j}+z \hat{k}=(1+2 \lambda) \hat{i}+(2+3 \lambda) \hat{j}+(-4+6 \lambda) \hat{k}\)
On comparing the coefficients of \(\hat{i}, \hat{j} \text { and } \hat{k}\), we get
\(\begin{aligned} x=1+2 \lambda, y=2+3 \lambda \text { and } z=-4+6 \lambda \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{x-1}{2}=\lambda, \frac{y-2}{3}=\lambda \text { and } \frac{z+4}{6}=\lambda \end{aligned}\)
\(\begin{aligned} \therefore \quad \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6} \end{aligned}\)
which is the required cartesian equation of a line.
Alternate Method
Let the equation of line passing through (1, 2, -4) is
\(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda_1\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)\)   ...(i)
Since, the line (i) is perpendicular to the given lines
\(\begin{aligned} \vec{r}=(8 \hat{i}-19 \hat{j}+10 \hat{k})+\lambda(3 \hat{i}-16 \hat{j}+7 \hat{k}) \end{aligned}\)
and  \(\begin{aligned} \vec{r}=(15 \hat{i}+29 \hat{j}+5 \hat{k})+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k}) \end{aligned}\)
Therefore, we have
\(\begin{aligned} \Rightarrow \quad\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right) \cdot(3 \hat{i}-16 \hat{j}+7 \hat{k})=0 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad 3 b_1-16 b_2+7 b_3=0 \end{aligned}\)                        ...(ii)
\(\begin{aligned} \text { and } \quad\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right) \cdot(3 \hat{i}+8 \hat{j}-5 \hat{k})=0 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad 3 b_1+8 b_2-5 b_3=0 \end{aligned}\)                         ...(iii)
[\(\because\) if two lines \(\vec{r}=\vec{a}_1+\lambda \vec{b}_1 \text { and } \vec{r}=\vec{a}_2+\lambda \vec{b}_2\) are perpendicular, then \(\vec{b}_1 \cdot \vec{b}_2=0 .\)]
Now, on solving Eqs. (ii) and (iii), we get
\(\begin{array}{rlrl} \frac{b_1}{80-56}= \frac{b_2}{21+15}=\frac{b_3}{24+48} \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \frac{b_1}{24} & =\frac{b_2}{36}=\frac{b_3}{72} \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \frac{b_1}{2} & =\frac{b_2}{3}=\frac{b_3}{6} \end{array}\) [multiplying by 12]
\(\Rightarrow\) b₁ = 2k, b₂ = 3k and b₃ = 6k, for some constant k.
Thus, the required vector equation of line is
\(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\),
where \(\lambda=\lambda_1 k\) is any constant.
Now, for cartesian equation do same as in above method.

We have, \(\vec{a}=\hat{i}+2 \hat{\jmath}+3 \hat{k} \text { and } \vec{b}=2 \hat{i}+4 \hat{\jmath}-5 \hat{k}\)
So, the diagonals of the parallelogram whose adjacent sides are \(\vec{a} \text { and } \vec{b}\) are given by
\(\begin{aligned} \vec{p} & =\vec{a}+\vec{b} \text { and } \vec{q}=\vec{a}-\vec{b} \end{aligned}\)
Now, \(\begin{aligned} \vec{p} & =(\hat{i}+2 \hat{j}+3 \hat{k})+(2 \hat{i}+4 \hat{j}-5 \hat{k}) \end{aligned}\)
\(\begin{aligned} =3 \hat{i}+6 \hat{j}-2 \hat{k} \end{aligned}\)
and \(\begin{aligned} \vec{q} & =(\hat{i}+2 \hat{j}+3 \hat{k})-(2 \hat{i}+4 \hat{j}-5 \hat{k}) \end{aligned}\)
\(\begin{aligned} =-\hat{i}-2 \hat{j}+8 \hat{k} \end{aligned}\)
\(\begin{aligned} \therefore \quad \hat{p} & =\frac{\vec{p}}{|\vec{p}|}=\frac{3 \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{9+36+4}} \end{aligned}\)
\(\begin{aligned} =\frac{3 \hat{i}+6 \hat{j}-2 \hat{k}}{7}=\frac{3}{7} \hat{i}+\frac{6}{7} \hat{j}-\frac{2}{7} \hat{k} \end{aligned}\)
and \(\begin{aligned} \hat{q} & =\frac{\vec{q}}{|\vec{q}|}=\frac{-\hat{i}-2 \hat{j}+8 \hat{k}}{\sqrt{1+4+64}}=\frac{-\hat{i}-2 \hat{j}+8 \hat{k}}{\sqrt{69}} \end{aligned}\)
\(\begin{aligned} =\frac{-1}{\sqrt{69}} \hat{i}-\frac{2}{\sqrt{69}} \hat{j}+\frac{8}{\sqrt{69}} \hat{k} \end{aligned}\)

We have,
\(\overrightarrow{A B}\) = (position vector of B) - (position vector of A)
\(=(\hat{i}-3 \hat{j}-5 \hat{k})-(2 \hat{i}-\hat{j}+\hat{k})=-\hat{i}-2 \hat{j}-6 \hat{k}\)
\(\begin{aligned} \overrightarrow{B C}=(3 \hat{i}-4 \hat{j}-4 \hat{k})-(\hat{i}-3 \hat{j}-5 \hat{k})=2 \hat{i}-\hat{j}+\hat{k} \end{aligned}\)
and \(\begin{aligned} \overrightarrow{C A}=(2 \hat{i}-\hat{j}+\hat{k})-(3 \hat{i}-4 \hat{j}-4 \hat{k}) \end{aligned}\)
\(=-\hat{i}+3 \hat{j}+5 \hat{k}\)
Here, \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=0\)
\(\Rightarrow\) A, B and C are the vertices of a triangle.
Now, \(\overrightarrow{B C} \cdot \overrightarrow{C A}=(2 \hat{i}-\hat{j}+\hat{k}) \cdot(-\hat{i}+3 \hat{j}+5 \hat{k})\)
= - 2 - 3 + 5 = 0
\(\Rightarrow \overrightarrow{B C} \perp \overrightarrow{C A} \Rightarrow \angle C=90^{\circ}\)

Now, area of \(\Delta A B C=\frac{1}{2}|\overrightarrow{C A} \times \overrightarrow{B C}|\)
\(=\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 5 \\ 2 & -1 & 1 \end{array}\right|=\frac{1}{2}|(8 \hat{i}-11 \hat{j}-5 \hat{k})|\)
\(=\frac{1}{2} \sqrt{64+121+25}\)
\(=\frac{1}{2} \sqrt{210}\) sq units

As \(\begin{vmatrix} 2x & -1 \\ 4 & 2 \end{vmatrix}=\begin{vmatrix} 3 & 0 \\ 2 & 1 \end{vmatrix}\)
⇒ 4x + 4 = 3 - 0
⇒ x = \(-\frac { 1 }{ 4 } \)

As 3 x \(\frac{k^2}{9}\) = 1 ⇒ k 土\(\sqrt3\)

\(\frac{1}{2}\left|\begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array}\right|=\pm 35\)

Sample space is given by {MFSD, MFDS, MSFD, MSDF, MDFS, MDSF, FMSD, FMDS, FSMD, FSDM, FDMS, FDSM, SFMD, SFDM, SMFD, SMDF, SDMF, SDFM DFMS, DFSM, DMSF, DMFS, DSMF, DSFM}, where F, M, D and S represent father, mother, daughter and son respectively.
∴ n(S) = 24
(i) (a): Let A denotes the event that daughter is at one end.
∴ n(A) = 12
and B denotes the event that father and mother are in the middle.
∴ n(B) = 4
Also, n(A \(\cap\) B) = 4
\(\therefore \ P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{4 / 24}{4 / 24}=1\)
(ii) (b): Let A denotes the event that mother is at right end.
∴ n(A) = 6
and B denotes the event that son and daughter are together.
∴ n(B) = 12
Also, n(A \(\cap\) B) = 4
\(\therefore \ P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{4 / 24}{12 / 24}=\frac{1}{3}\)
(iii) (c) : Let A denotes the event that father and mother are in the middle.
∴ n(A) = 4
and B denotes the event that son is at right end.
∴ n(B) = 6
Also, n(A \(\cap\) B) = 2
\(\therefore \ P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{2 / 24}{6 / 24}=\frac{1}{3}\)
(iv) (d): Let A denotes the event that father and son are standing tbgether.
∴ n(A) = 12
and B denotes the event that mother and daughter are standing together.
∴ n(B) = 12
Also, n(A \(\cap\) B) = 8
\(\therefore \ P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{8 / 24}{12 / 24}=\frac{2}{3}\)
(v) (a): Let A denotes the event that father and mother are on either of the ends.
∴ n(A) = 4
and B denotes the event that son is at second position from the right end.
∴ n(B) = 6
Also, n(A \(\cap\) B) = 2
\(\therefore \ P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{2 / 24}{6 / 24}=\frac{1}{3}\)

(i) (d) : Clearly, the coordinates of A are (8,10, 0) and Dare (0, 0, 30)
∴ Equation of AD is given by
\(\frac{x-0}{8-0}=\frac{y-0}{10-0}=\frac{z-30}{-30} \)
\(\Rightarrow \quad \frac{x}{4}=\frac{y}{5}=\frac{30-z}{15}\)
(ii) (b): The coordinates of point Care (15, -20, 0) and Dare (0, 0, 30)
∴ Length of the cable DC
\(=\sqrt{(0-15)^{2}+(0+20)^{2}+(30-0)^{2}}\)
\(=\sqrt{225+400+900}=\sqrt{1525}=5 \sqrt{61} \mathrm{~m}\)
(iii) (a) : Since, the coordinates of point Bare (-6, 4, 0) and Dare (0, 0, 30), therefore vector DB is
\((-6-0) \hat{i}+(4-0) \hat{j}+(0-30) \hat{k}, \text { i.e., }-6 \hat{i}+4 \hat{j}-30 \hat{k}\)
(iv) (b) : Required sum
\(=(\hat{i}+10 \hat{j}-30 \hat{k})+(-6 \hat{i}+4 \hat{j}-30 \hat{k})+(15 \hat{i}-20 \hat{j}-30 \hat{k}) \)
\(=17 \hat{i}-6 \hat{i}-90 \hat{k}\)
(v) (a): Clearly, OA = \(\sqrt{8^{2}+10^{2}}=\sqrt{164}\)
\(O B=\sqrt{6^{2}+4^{2}}=\sqrt{36+16}=\sqrt{52} \ \text { and } O C=\sqrt{15^{2}+20^{2}}=\sqrt{225+400}=\sqrt{625} \)

Let the cost of 1 pen = Rs. x, the cost of 1 bag = Rs. y, and the cost of 1 instrument box = Rs. z
According to the question, we have
5x + 3y + z = 16, 2x + Y + 3z = 19, x + 2y + 4z = 25
This system of equation can be written as AX = B,
where \(A=\left[\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right], B=\left[\begin{array}{l} 16 \\ 19 \\ 25 \end{array}\right] \) and \(X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\) 
IAI = 5(4 - 6) - 3(8 - 3) + 1(4 - 1)
\(=-10-3(5)+3=-22 \neq 0\)
\(\therefore\)  A ^-1 exists.
Now, X = A^-1B, where \(A^{-1}=\frac{1}{|A|} \operatorname{adj} A\) 
Here, \(\operatorname{adj} A=\left[\begin{array}{ccc} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -13 & -1 \end{array}\right]^{\prime}=\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\) 
\(\therefore \quad A^{-1}=\frac{1}{-22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\) 
\(\therefore \quad X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{-22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\left[\begin{array}{c} 16 \\ 19 \\ 25 \end{array}\right]\) 
\(=\frac{1}{-22}\left[\begin{array}{c} -32-190+200 \\ -80+361-325 \\ 48-133-25 \end{array}\right]=\frac{-1}{22}\left[\begin{array}{c} -22 \\ -44 \\ -110 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 5 \end{array}\right]\) 
\(\therefore\) x = 1, y = 2, z = 5
Hence, cost of one pen, one bag and an instrument box isRs. 1, Rs. 2 and Rs. 5 respectively.
(i) (c) : Cost of one pen is Rs. 1.
(ii) (a) : Cost of one pen. and one bag = Rs. (1 + 2) = Rs. 3
(iii) (b) : Cost of one pen and one instrument box
= Rs. (1 + 5) = Rs. 6
(iv) (c) : According to the definition of determinant, determinartt is a number associated to a square matrix.
(v) (b) : Given matrix equation is AB = AC Pre-multiplying by A^-I on both sides, we get
\(A^{-1} A B=A^{-1} A C \Rightarrow\left(A^{-1} A\right) B=\left(A^{-1} A\right) C\) 
\(\Rightarrow \quad I B=I C \quad\left(\because A A^{-1}=A^{-1} A=I\right)\) 
\(\Rightarrow B=C\) 
Since A^-1 exists only if A i's non-singular
\(\therefore\) For B = C, A should be non-singular

(d) Assertion is incorrect, Reason is correct.

(b) Both A and R are correct; R is not the correct explanation of A