Here, \(\vec { u } =\hat { i } +2\hat { j } +3\hat { k } ,\vec { n } =\hat { i } +2\hat { j } +4\hat { k } \), p = 38. Then the position vector of the image \(\vec { v } \)of
\(\vec { u } =\hat { i } +2\hat { j } +3\hat { k } \) is given by \(\vec { v } =\vec { u } +\frac { 2[p-(\vec { u } .\vec { n } )] }{ { \left| \vec { n } \right| }^{ 2 } } \vec { n } \)
\(\vec { v } =(\hat { i } +2\hat { j } +3\hat { k } )+\frac { 2[38-(\hat { i } +2\hat { j } +3\hat { k } ).(\hat { i } +2\hat { j } +4\hat { k } ))] }{ (\hat { i } +2\hat { j } +4\hat { k } ).(\hat { i } +2\hat { j } +4\hat { k } ) } (\hat { i } +2\hat { j } +4\hat { k } )\)
That is \(\vec { v } =(\hat { i } +2\hat { j } +3\hat { k } )+2(\frac { [38- 17]}{21}) (\hat { i } +2\hat { j } +4\hat { k } )= (3\hat { i } +6\hat { j } +11\hat { k } )\)
Therefore, the image of the point with position vector \(\hat { i } +2\hat { j } +3\hat { k } \) is \(3\hat { i } +6\hat { j } +11\hat { k } \)

We know that the vector equation of a plane passing through the line of intersection of the planes
\(\vec { r } .\vec { { n }_{ 1 } } ={ d }_{ 1 }\) and \(\vec { r } .\vec { { n }_{ 2 } } ={ d }_{ 2 }\) is given by \((\vec { r } .\vec { { n }_{ 1 } } -{ d }_{ 1 })+\lambda (\vec { r } .\vec { { n }_{ 2 } } -{ d }_{ 2 })=0\)
Substituting \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } ,\vec { { n }_{ 1 } } =\hat { i } +\hat { j } +\hat { k } ,\vec { { n }_{ 2 } } =2\hat { i } -3\hat { j } +5\hat { k } \), \({ d }_{ 1 }=1,{ d }_{ 2 }=-2\) in the above equation, we get
(x + y + z + 1) + \(\lambda \) (2x - 3y + 5z - 2) = 0
Since this plane passes through the point (−1, 2,1) , we get λ = \(\frac{3}{5}\), and hence the required equation
of the plane is 11x−4y+20z=1 .

Given equation of plane is x + 2y + 3z = 2
Length of perpendicular from (4, 3, 2) to the plane is
\(d=\cfrac { 4+2\left( 3 \right) +3\left( 2 \right) }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } } } =\cfrac { 4+6+6 }{ \sqrt { 14 } } \)
= \(\cfrac { 14 }{ \sqrt { 14 } } =\cfrac { \sqrt { 14 } .\sqrt { 14 } }{ \sqrt { 14 } } =\sqrt { 14 } \) units
Let us find the image of the point (4,3,2) to the plane x + 2y + 3z = 2
Here \(\vec { u } =4\hat { i } +3\hat { j } +2\hat { k } ,\vec { n } =\hat { i } +2\hat { j } +3\hat { k } \)
Then the image \(\vec { v } =\vec { u } +\cfrac { 2\left[ p-\left( \vec { u } .\vec { n } \right) \right] }{ \left| \vec { n } \right| ^{ 2 } } \)
\(\vec { v } =\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) +\cfrac { 2\left[ 2-\left( 4+6+6 \right) \right] }{ \left( \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } } \right) } \left( \hat { i } +2\hat { j } +3\hat { k } \right) \)
= \(\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) +\frac { 2\left( 2-16 \right) }{ 14 } \left( \hat { i } +2\hat { j } +3\hat { k } \right) \)
= \(\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) +\cfrac { 2\left( -14 \right) }{ 14 } \left( \hat { i } +2\hat { j } +3\hat { k } \right) \)
= \(\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) -2\left( \hat { i } +2\hat { j } +3\hat { k } \right) \)
= \(\left( 4\hat { i } +3\hat { j } +2\hat { k } \right) -2\left( \hat { i } +4\hat { j } -6\hat { k } \right) \)
= \(2\hat { i } -\hat { j } -4\hat { k } \)
\(\therefore\) The foot of the \(\bot \) from (4, 3, 2) to the plane is
\(\cfrac { \left( 4\hat { i } +3\hat { j } +2\hat { k } \right) +\left( 2\hat { i } -\hat { j } -4\hat { k } \right) }{ 2 } \)
= \(\cfrac { 6\hat { i } +2\hat { j } -2\hat { k } }{ 2 } =3\hat { i } +\hat { j } -\hat { k } \)
Hence, the co-ordinates of the foot of the perpendicular is (3, 1, -1)

Given equation of line is
\(\frac { x-1 }{ 1 } =\frac { y }{ 2 } =z+1=s\)
\(\Rightarrow x-1=s\Rightarrow x=s+1\)
\(\frac { y }{ 2 } =s\Rightarrow y=2s\)
\(z+1=s\Rightarrow z=s-1\)
\(\therefore\)  Any point on the line is of the form (s+1, 2s, s-1)
This point lies on the plane 2x-y+2z = 2
\(\Rightarrow 2(s+1)-(2s)+2(s-1)=2\)

\(\\ \Rightarrow 2s=2\Rightarrow s=1\)
when s = 1, the point is (1 + 1, 2(1), 1 - 1)
= (2, 2, 0)
\(sin\ \theta = \frac{|\vec b.\vec n|}{|\vec b||\vec n|} \)
\(\vec b = \vec i+2\vec j+\vec k, \quad \vec n = 2\vec i-\vec j+ 2\vec k \)
\(\vec b.\vec n = 2-2+2 = 2 \)
\(|\vec b|= \sqrt{1+4+1 }=\sqrt 6 \)
\( |\vec n|=\sqrt{ 4+1+4}= \sqrt 9 = 3 \)
\( sin\ \theta = \frac{|2|}{\sqrt 6 \times 3} = \frac {2}{3\sqrt 6} \)
\(\theta = sin ^{-1}(\frac{2}{3\sqrt 6})\)
which is the point of intersection of the plane and the line

Comparing the two given lines with
\(\vec { r } =\vec { a } +t\vec { b } ,\vec { r } =\vec { c } +s\vec { d } \)
we have, \(\vec { a } =-\hat { i } -3\hat { j } -5\hat { k } ,\vec { b } =3\hat { i } +5\hat { j } +7\hat { k } ,\vec { c } =2\hat { i } +4\hat { j } +6\hat { k } \)  and \(\vec { d } =\hat { i } +4\hat { j } +7\hat { k } \)
We know that the two given lines are coplar, if \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\) = 0
Here, \(\vec { b } \times \vec { d } \left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{matrix} \right| =7\hat { i } -14\hat { j } +7\hat { k } \)  and  \(\vec { c } -\vec { a } =3\hat { i } +7\hat { j } +11\hat { k } \)
Then, \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=(3\hat { i } +7\hat { j } +11\hat { k } )(7\hat { i } -14\hat { j } +7\hat { k } )=0\)
Therefore the two given lines are coplanar. Then we find the non parametric form of vector equation of the plane containing the two given coplanar lines. We know that the plane containing the two given coplanar lines is
\((\vec { r } -\vec { a } ).(\vec { b } \times \vec { d } )\)= 0
which implies that \((\vec { r } -(-\hat { i } -3\hat { j } -5\hat { k } )).(7\hat { i } -14\hat { j } +7\hat { k } )\) = 0. 
Thus, the required non-parametric vector equation of the plane containing the two given coplanar lines is
\(\vec { r } .(\hat { i } -2\hat { j } +\hat { k } )\) = 0.

\(\vec { r } =(5\hat { i } +7\hat { j } -3\hat { k } )+s(-4\hat { i } +4\hat { j } -5\hat { k } )\) and \(\vec { r } =(8\hat { i } +4\hat { j } +5\hat { k } )+t(7\hat { i } +\hat { j } +3\hat { k } )\)
Let \(\vec { a } =5\hat { i } +7\hat { j } -3\hat { k } ,\vec { b } =4\hat { i } +4\hat { j } -5\hat { k } \)
\(\vec { c } =8\hat { i } +4\hat { j } +5\hat { k } \ and\ \vec { d } =7\hat { i } +\hat { j } +3\hat { k } \)
We know that the two given lines are co-planar
if \(\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =0\)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & 4 & -5 \\ 7 & 1 & 3 \end{matrix} \right| \)
= \(\hat { i } \left( 12+5 \right) -\hat { j } \left( 12+35 \right) +\hat { k } \left( 4-28 \right) \)
= \(17\hat { i } -47\hat { j } -24\hat { k } \)
\(\left( \vec { c } -\vec { a } \right) =\left( 8-5 \right) \hat { i } +\left( 4-7 \right) \hat { j } +\left( 5+3 \right) \hat { k } \)
= \(3\hat { i } -3\hat { j } +8\hat { k } \)
Now,\(\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =\left( 3\hat { i } -3\hat { j } +8\hat { k } \right) \)
\(\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) \)
= \(51+141-192=192-192=0\)
\(\therefore\) The two given lines are co-planar.
The plane containing the two given co-planar lines is
\(\left( \vec { r } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =0\)
\(\Rightarrow \left( \vec { r } -5\hat { i } +7\hat { j } -3\hat { k } \right) \times \left( 17\hat { i } -47\hat { j } -24\hat { k } \right) =0\)
\(\Rightarrow \vec { r } .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) -\left[ \left( 5\hat { i } +7\hat { j } -3\hat { k } \right) .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) \right] =0\)
\(\Rightarrow \vec { r } .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) =\left[ 85-329+72 \right] \)
\(\Rightarrow \vec { r } .\left( 17\hat { i } -47\hat { j } -24\hat { k } \right) =-172 \)
which is the required vector equation of the plane.

The plane containing the line
\(\vec { r } =(\hat { i } -\hat { j } +3\hat { k } )+t(2\hat { i } -\hat { j } +4\hat { k } )\)
∴ The required plane is passing through the point \(\vec { a } =\hat { i } -\hat { j } +3\hat { k } \) and parallel to a vector \(\vec { b } =2\hat { i } -\hat { j } +4\hat { k } \) Also, the plane is perpendicular to the plane 
\(\vec { c } =\hat { i } +2\hat { j } +\hat { k } \) 
∴ The parametric form of vector equation of the plane passing through one point (\(\vec { a } \)) and parallel to two vectors \(\vec { b } \) and \(\vec { c } \) 
\(\vec { r } =\vec { a } +s\vec { b } +t\vec { c } \) where s, t ∈ R
⇒ \(\vec { r } .(\hat { i } -\hat { j } +3\hat { k } )+s(2\hat { i } -\hat { j } +4\hat { k } )+t(\hat { i } +2\hat { j } +\hat { k } )\) s, t  ∈ R 
Cartesian equation is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 & y+1 & z-3 \\ 2 & -1 & 4 \\ 1 & 2 & 1 \end{matrix} \right| =0\)
⇒ (x - 1)(-1 - 8) - (y + 1)(2 - 4) + (z - 3)(4 + 1) = 0
⇒ (x - 1)(-9) - (y + 1)(-2) + (z - 3)5 = 0
⇒ -9x + 9 +2y + 2 + 5z - 15 = 0
⇒ -9x + 2y + 5z - 4 = 0
⇒ 9x - 2y - 5z + 4 = 0

Given plane is passing through the points
\(\vec { a } =2\hat { i } +2\hat { j } +2\hat { k }, \vec { b } =9\hat { i } +3\hat { j } +6\hat { k } \)
Equation of the given plane is 2x + 6y + 6z = 9. It can be written as \(\vec { r } .(2\hat { i } +6\hat { j } +6\hat { k } )=9\)
Since the given plane is perpendicular to \(2\hat { i } +6\hat { j } +6\hat { k } \), the required plane is parallel to \(\vec { c } =2\hat { i } +6\hat { j } +6\hat { k } \). Hence, parametric form of vector equation of plane passing through two points and parallel to a vector is
\(\vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t\vec { c } ,s,t\in R\)
\(\vec { r } =2\hat { i } +2\hat { j } +\hat { k } +s(7\hat { i } +\hat { j } +5\hat { k } )+t(2\hat { i } +6\hat { j } +6\hat { k } ),s,t\in R\)
Cartesian equation of the plane is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-2 & y-2 & z-1 \\ 7 & 1 & 5 \\ 2 & 6 & 6 \end{matrix} \right| =0\)
⇒ (x-2)(6-30) - (y-2)(42-10) + (z-1)(42-2) = 0
⇒ (x - 2)(-24) - (y - 2)(32) + (z - 1)(40) = 0
⇒ 24x + 48 - 32y + 64 + 40z - 40 = 0
⇒ -24x - 32y + 40z + 72 = 0
\(\div\) - 8 we get
3x+ 4y - 5z - 9 = 0 is the Cartesian form.
∴ The parametric form of vector equation is
\(\vec { r } =\vec { r } (3\vec { i } +4\vec { j } -5\vec { k } )=9\)

The plane passes through the point.
\(\vec { a } =2\hat { i } +3\hat { j } +6\hat { k } \) and parallel to the lines \(\frac{x-1}{2}\)
\(=\frac { y+1 }{ 3 } =\frac { z-3 }{ 1 } and\frac { x+3 }{ 2 } =\frac { y-3 }{ -5 } =\frac { z+1 }{ -3 } \)
\(\Rightarrow \vec { b } =2\hat { i } +3\hat { j } +\hat { k }\ and\ \vec { c } =2\hat { i } -5\hat { j } -3\hat { k } \)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 1 \\ 2 & -5 & -3 \end{matrix} \right| \)
\(=\hat { i } (-9+5)-\hat { j } (-6-2)+\hat { k } (-10-6)\)
\(=-4\hat { i } +8\hat { j } -16\hat { k } \)
The non-parametric vector equation of the plane is
\((\vec { r } .\vec { a } ).(\vec { b } \times \vec { c } )=0,\)
\(\Rightarrow [\vec { r } (2\hat { i } +3\hat { j } +16\hat { k } ).(-4\hat { i } +8\hat { j } -16\hat { k } )]=0\)
\(\Rightarrow [\vec { r } .(-4\hat { i } +8\hat { j } -16\hat { k } )]-(-8+24-96)=0\)
\(\Rightarrow \vec { r } .(-4\hat { i } +8\hat { j } -16\hat { k } )=-80\)
\(\div -4,\) We get
\(\vec { r } .(\hat { i } -2\hat { j } +4\hat { k } )=20\)
\(Let\quad \vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\Rightarrow (x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } -2\hat { j } +4\hat { k } )=20\)
\(\Rightarrow x=2y+4z=20\)
\(\Rightarrow x-2y+4z-20=0\)

We observe that the required plane is parallel to the vectors \(\vec { b } =2\hat { i } +3\hat { j } +6\hat { k } \), \(\vec { c } =\hat { i } +\hat { j } -\hat { k } \) and passing through the point(0,1,-5) with position vector \(\vec { a } \). We observe that \(\vec { b } \) is not parallel to \(\vec { c } \).
Then the vector equation of the plane in non-parametric form is given by \((\vec { r } -\vec { a } ).(\vec { b } \times \vec { c } )=0\) ..(1)
Substituting \(\vec { a } =\hat { j } -5\hat { k } \) and \(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{matrix} \right| =-9\hat { i } +8\hat { j } -\hat { k } \) in equation (1), we get
\((\vec { r } (-\hat { j } -5\hat { k } )).(-9\hat { i } +8\hat { j } -\hat { k } )=0\), which implies that
\(\vec { r } (-9\hat { i } +8\hat { j } -\hat { k } )=13\)
\(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) is the position vector of an arbitrary point on the plane, then from the above equation, we get the Cartesian equation of the plane as -9x + 8y - z = 13 or 9x - 8y + z + 13 = 0