(i) (b): The line along which motorcycle A is running, \(\vec{r}=\lambda(\hat{i}+2 \hat{j}-\hat{k})\) is which can be rewritten as \((x \hat{i}+y \hat{j}+z \hat{k})=\lambda \hat{i}+2 \lambda \hat{j}-\lambda \hat{k}\)
\(\Rightarrow x=\lambda, y=2 \lambda, z=-\lambda \Rightarrow \frac{x}{1}=\lambda, \frac{y}{2}=\lambda, \frac{z}{-1}=\lambda\)
Thus, the required cartesian equation is \(\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}\)
(ii) (d): Clearly, D.R:s of the required line are <  1, 2, -1  >
∴ D.Cs are \( <\frac{1}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}, \frac{2}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}, \frac{-1}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}> \)
\(\text { i.e., }<\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}>\)
(iii) (d): The line along which motorcycle B is running, is \(\vec{r}=(3 \hat{i}+3 \hat{j})+\mu(2 \hat{i}+\hat{j}+\hat{k})\), which is parallel to the vector \(2 \hat{i}+\hat{j}+\hat{k}\).
∴ D.R.'s of the required line are < 2, 1, 1 >.
(iv) (d): Here, \(\vec{a}_{1}=0 \hat{i}+0 \hat{j}+0 \hat{k}, \vec{a}_{2}=3 \hat{i}+3 \hat{j}, \vec{b}_{1}=\hat{i}+2 \hat{j}-\hat{k} \vec{b}_{2}=2 \hat{i}+\hat{j}+\hat{k}\)
\(\therefore \vec{a}_{2}-\vec{a}_{1}=3 \hat{i}+3 \hat{j}\)
and  \(\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 1 & 1 \end{array}\right|=3 \hat{i}-3 \hat{j}-3 \hat{k}\) 
Now, \(\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)=(3 \hat{i}+3 \hat{j}) \cdot(3 \hat{i}-3 \hat{j}-3 \hat{k})\)
= 9 - 9 = 0.
Hence, shortest distance between the given lines is 0.
(v) (c): Since, the point (1, 2, -1) satisfy both the equations of lines, therefore point of intersection of given lines is (1, 2, -1). So, the motorcycles will meet with an accident at the point (1, 2, -1).

(i) (b): Clearly, the coordinates of A are (8, -6, 0) and that of E are (0, 0, 24).
Also, cartesian equation of line along EA is given by
\(\frac{x-0}{8-0}=\frac{y-0}{-6-0}=\frac{z-24}{0-24}\)
\(\Rightarrow \frac{x}{8}=\frac{y}{-6}=\frac{z-24}{-24} \Rightarrow \frac{x}{-4}=\frac{y}{3}=\frac{z-24}{12}\)
(ii) (c): Clearly, the coordinates of Dare (-8, -6, 0) and that of E are (0, 0, 24)
\(\therefore \text { Vector } \overline{E D} \text { is }(-8-0) \hat{i}+(-6-0) \hat{j}+(0-24) \hat{k} \ \text { i.e., }-8 \hat{i}-6 \hat{i}-24 \hat{k} \)
(iii) (b): Since, the coordinates of Bare (8, 6, 0) and that of E are (0, 0, 24), therefore length of cable
\(E B=\sqrt{(8-0)^{2}+(6-0)^{2}+(0-24)^{2}} \)
\(=\sqrt{64+36+576}=\sqrt{676}=26 \text { units }\)
(iv) (d): Since, the coordinates of Care (-8,6,0) therefore length of cable EC = \(\sqrt{(-8-0)^{2}+(6-0)^{2}+(0-24)^{2}}\)
\(=\sqrt{64+36+576}=\sqrt{676}=26 \mathrm{units}\)
Similarly, length of cable EA = ED = 26 units.
(v) (c): Sum of all vectors along the cables
\(=\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overline{E D}\)
\( =(8 \hat{i}-6 \hat{j}-24 \hat{k})+(8 \hat{i}+6 \hat{j}-24 \hat{k})+(-8 \hat{i}+6 \hat{j}-24 \hat{k})+(-8 \hat{i}-6 \hat{j}-24 \hat{k}) \)
\(=-96 \hat{k}\)

(i) (b) : Equation of plane is x - 2y + 2z = 3
∴ D.R's of normal to the plane are < 1, - 2, 2 >, which is also the D.R.'s of perpendicular from the point (3, -2, 1) to the given plane.
(ii) (c) : Required length = Perpendicular distance from (3, -2, 1) to the plane x - 2y + 2z = 3
\(=\left|\frac{3-2(-2)+2(1)-3}{\sqrt{1^{2}+(-2)^{2}+2^{2}}}\right|=\frac{6}{3}=2 \text { units }\)
(iii) (b) : The equation of perpendicular from the point (x₁, y₁, z₁) to the plane ax + by + cz = d is given by
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
Here, x₁ = 3, y₁ = -2,  z₁= 1 and a = 1, b = -2, c = 2
∴ Required equation is \(\frac{x-3}{1}=\frac{y+2}{-2}=\frac{z-1}{2}\)
(iv) (d) : The equation of the plane parallel to the plane x - 2y + 2z - 3 = 0 is x - 2y + 2z + \(\lambda\) = 0
Now, distance of this plane from the point (3, -2, 1) is
\(\left|\frac{3+4+2+\lambda}{\sqrt{1^{2}+(-2)^{2}+2^{2}}}\right|=\left|\frac{9+\lambda}{3}\right|\)
But, this distance is given to be unity
\(\therefore|9+\lambda|=3 \Rightarrow \lambda+9=\pm 3 \Rightarrow \lambda=-6 \text { or }-12\)
Thus, required equation of planes are
x - 2y + 2z - 6 = 0 or x- 2y + 2z - 12 = 0
(v) (a): Let the coordinate of image of (3, -2, 1) be Q(r + 3, -2r - 2, 2r + 1)
Let R be the mid-point of PQ, then coordinate of R be \(\left(\frac{r+6}{2}, \frac{-2 r-4}{2}, r+1\right)\)
Since, R lies on the plane x - 2y + 2z = 3
\( \therefore \quad\left(\frac{r+6}{2}\right)-2\left(\frac{-2 r-4}{2}\right)+2(r+1)=3 \)
\(\Rightarrow \quad 9 r=-12 \Rightarrow r=-\frac{4}{3}\)
Thus, the coordinates of Q be \(\left(\frac{5}{3}, \frac{2}{3}, \frac{-5}{3}\right)\)

(i) (b) : D.R:s of OA are < 1-0, 0-0, 0-0 >, i.e., < 1, 0, 0 >.
(ii) (a) : Equation of diagonal OB' is \(\frac{x-0}{1}=\frac{y-0}{2}=\frac{z-0}{3} \text { i.e., } \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)
(iii) (c) : OABC is xy-plane, therefore its equation is z = 0.
(iv) (c) : Plane O'A'B'C is parallel to xy-plane passing through (0, 0, 3), therefore its equation is z = 3.
(v) (a) : Plane ABB' A' is parallel to yz-plane passing through (1, 0, 0), therefore its equation is x = 1.

(i) (a): Eliminating 't' from the given equations, we get equation of path as, \(\frac{x}{2}=\frac{y}{-4}=\frac{z}{4} \text { or } \frac{x}{1}=\frac{y}{-2}=\frac{z}{2}\).
Thus, the path of the rocket represents a straight line.
(ii).(b) : Since, only (1, -2, 2) satisfy the equation of path of rocket therefore (1, -2, 2) lie on the path of rocket.
(iii) (b): For t = 10 sec, we have x = 20,y = -40, z = 40 
Now, required distance = \(\sqrt{x^{2}+y^{2}+z^{2}}\)
\(=\sqrt{20^{2}+(-40)^{2}+(40)^{2}}=\sqrt{400+1600+1600}\)
\(=\sqrt{3600}=60 \mathrm{~km}\)
(iv) (d) : Clearly, height of rocket from the ground = z-coordinate of given position = 6 km
(v) (b) : Let Q be the image of point P(1, -2, 2) in the plane 3x - y + 4z = 2. Then, equation of PQ is
\(\frac{x-1}{3}=\frac{y+2}{-1}=\frac{z-2}{4}\)
Let the coordinates of Q be (3r +1, -r - 2, 4r + 2).
Let R be the mid -point of PQ. Then, coordinates of R are
\(\left(\frac{3 r+2}{2}, \frac{-r-4}{2}, \frac{4 r+4}{2}\right) \text { or }\left(\frac{3}{2} r+1, \frac{-r}{2}-2,2 r+2\right)\)
Since, R lies on 3x - y + 4z = 2.
\( \therefore \quad 3\left(\frac{3}{2} r+1\right)-\left(\frac{-r}{2}-2\right)+4(2 r+2)=2 \)
\(\Rightarrow \frac{9 r}{2}+3+\frac{r}{2}+2+8 r+8=2 \\ \)
\(\Rightarrow 13 r+13=2 \Rightarrow r=\frac{-11}{13}\)
Hence, the coordinates of Q are
\(\left(\frac{-33}{13}+1, \frac{11}{13}-2, \frac{-44}{13}+2\right) \text { i.e., }\left(\frac{-20}{13}, \frac{-15}{13}, \frac{-18}{13}\right)\)