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 \(3\hat { i } +6\hat { j } +11\hat { k } \)

\(\frac { x-1 }{ 2 } =\frac { y+1 }{ \lambda } =\frac { z }{ 2 } \) and \(\frac { x-1 }{ 2 } =\frac { y+1 }{ \lambda } =\frac { z }{ \lambda } \)
\(\therefore \vec { a } =\hat { i } -\hat { j } ,\vec { b } =2\hat { i } +\lambda \hat { j } +2\hat { k } \)
\(\vec { c } =-\hat { i } -\hat { j } ,\vec { d } =5\hat { i } +2\hat { j } +\lambda \vec { k } \)
\(\left( \vec { c } -\vec { a } \right) =-2\hat { i } ,\)
and \(\left( \vec { b } \times \vec { d } \right) =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & \lambda & 2 \\ 5 & 2 & \lambda \end{matrix} \right| \)
= \(\hat { i } \left( { \lambda }^{ 2 }-4 \right) -\hat { j } \left( 2\lambda -10 \right) +\hat { k } \left( 4-5\lambda \right) \)
Since the given lines are co-planar,
\(\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =0\)
\(\Rightarrow \left( -2\hat { i } \right) .\left[ \left( { \lambda }^{ 2 }-4 \right) \hat { i } -\hat { j } \left( 2\lambda -10 \right) +\hat { k } \left( 4-5\lambda \right) \right] =0\)
\(\Rightarrow -2\left( { \lambda }^{ 2 }-4 \right) =0\)
\(\Rightarrow { \lambda }^{ 2 }=4\)  \(\left[ \because -2\neq 0 \right] \)
\(\Rightarrow \lambda =\pm \sqrt { 4 } =\pm 2\)
The Cartesian equation of the plane containing the given lines is
\(\left| \begin{matrix} x-{ x }_{ 2 } & y-{ y }_{ 2 } & z-{ z }_{ 2 } \\ { b }_{ 1 } & { b }_{ 2 } & b_{ 3 } \\ { d }_{ 1 } & { d }_{ 2 } & { d }_{ 3 } \end{matrix} \right| \)
\(\Rightarrow \left| \begin{matrix} x+1 & y+1 & z \\ 2 & 2 & 2 \\ 5 & 2 & 2 \end{matrix} \right| =0\left[ \because \lambda =2 \right] \)
\(\Rightarrow \left( x+1 \right) \left( 4-4 \right) -\left( y+1 \right) \left( 4-10 \right) +z\left( 4-10 \right) =0\)
\(\Rightarrow \left( x+1 \right) \left( 0 \right) -\left( y+1 \right) \left( -6 \right) +z\left( -6 \right) =0\)
\(\Rightarrow 6\left( y+1 \right) -6z=0\)
\(\Rightarrow y+1-z=0\)
\(\Rightarrow y+z+1=0\) which is the required equation of the plane containing the given lines

Gives
\(\frac { x-2 }{ 1 } =\frac { y-3 }{ 1 } =\frac { z-4 }{ 3 } \) and \(\frac { x-1 }{ -3 } =\frac { y-4 }{ 2 } =\frac { z-5 }{ 1 } \)
\(\therefore \vec { a } =-2\hat { i } -3\hat { j } -4\hat { k } ,\vec { b } =\hat { i } +\hat { j } +3\hat { k } \)
\(\vec { c } =-\hat { i } -4\hat { j } -5\hat { k } ,\vec { d } =-3\hat { i } +2\hat { j } +\hat { k } \)
The two given lines are co-planar
\(y\left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) \)
\(\left( \vec { c } -\vec { a } \right) =-\hat { i } +\hat { j } +\hat { k } \)
\(\vec { b } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 1 & 3 \\ -3 & 2 & 1 \end{matrix} \right| =\sqrt { 2 } \)
= \(\hat { i } \left( 1-6 \right) -\hat { j } (1+9)+\hat { k } \left( 2+3 \right) \)
= \(-5\hat { i } -10\hat { j } +5\hat { k } \)
\(\therefore \left( \vec { c } -\vec { a } \right) .\left( \vec { b } \times \vec { d } \right) =\left( -\hat { i } +\hat { j } +\hat { k } \right) .\left( -5\hat { i } -10\hat { j } +5\hat { k } \right) \)
= 5-10 + 5= 10-10 = 0
Hence, the given lines are co-planar. Its Cartesian equation is
\(\left| \begin{matrix} x-{ x }_{ 2 } & y-{ y }_{ 2 } & z-{ z }_{ 2 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { d }_{ 1 } & { d }_{ 2 } & { d }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 & y-4 & z-5 \\ 1 & 1 & 3 \\ -3 & 2 & 1 \end{matrix} \right| =0\)
\(\Rightarrow \left( x-1 \right) \left( 1-6 \right) -\left( y-6 \right) \left( 1+9 \right) +\left( z-5 \right) \left( 2+3 \right) =0\)
\(\Rightarrow \left( x-1 \right) \left( -5 \right) -\left( y-5 \right) \left( 10 \right) +\left( z-5 \right) \left( 5 \right) =0\)
\(\Rightarrow -5x+5-10y+40+5z-25=0\)
\(\Rightarrow -5x-10y+5z+20=0\)
\(\div\) -5, we get
x + 2y - z - 4 = 0 which is the equation of the plane containing the given lines

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

Equation of the plane passing through the point
\(=\vec { a } =\hat { i } -2\hat { j } +4 \) .......(1)
Equation of the given plane is x + 2y - 3z = 11
\(\Rightarrow \vec { r } .(\hat { i } +2\hat { j } -3\hat { k } )=11\)
The given plane is perpendicular to the vector \(\hat { i } +2\hat { j } -3\hat { k } \)
∴ The required plane is parallel to the vector 
\(\vec { b } =\hat { i } +2\hat { j } -3\hat { k } \).......(2)
The given plane is parallel to the line 
\(\frac { x+7 }{ 3 } =\frac { y+3 }{ -1 } =\frac { z }{ 1 } \) Whose direction ratios are  3,-1,1
The required plane is parallel to the vector
 \(\vec { c } =3\hat { i } -\hat { j } +\hat { k } \)     (3)
∴  The non-parametric vector equation of the plane passing through a point (\(\vec { a } \)) and parallel to two vectors \(\vec { b } \) and \(\vec { c } \) is
\((\vec { r } -\vec { a } ).(\vec { b } \times \vec { c } )=0\)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & -3 \\ 3 & -1 & 1 \end{matrix} \right| \)
\(=\hat { i } (2-3)-\hat { j } (1+9)+\hat { k } (-1-6)\)
\(=-\hat { i } -10\hat { j } -7\hat { k } \)
\(\therefore (\vec { r } -(\hat { i } -2\hat { j } +4\hat { k } )).(-\hat { i } -10\hat { j } -7\hat { k } )=0\)
\([\vec { r } -(-\hat { i } -10\hat { j } -7\hat { k } )]-[(\hat { i } -2\hat { j } +4\hat { k } ).(-\hat { i } -10\hat { j } -7\hat { k } )]=0\)
\(\Rightarrow \vec { r } .(-\hat { i } -10\hat { j } -7\hat { k } )-[-1+20-28]=0\)
\(\Rightarrow \vec { r } .(-\hat { i } -10\hat { j } -7\hat { k } )\times 9=0\)
\(\Rightarrow \vec { r } .(\hat { i } +10\hat { j } +7\hat { k } )-9=0\)
\(\Rightarrow \vec { r } .(\hat { i } +10\hat { j } +7\hat { k } )=9\)
Let \(\Rightarrow \vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\therefore (x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } +10\hat { j } +7\hat { k } )=9\)
⇒ x+10y+7z = 9 which is the required Cartesian equation of the plane.