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.

The plane passes through two points
\(\vec { a } =2\hat { i } +2\hat { j } +\hat { k }\ and\ \vec { b } =\hat { i } -2\hat { j } +3\hat { k } \)
The straight line passing through the points
(2, 1, -3) and (-1, 5, -8) is
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } .\)
\(\Rightarrow \frac { x-2 }{ -1-2 } =\frac { y-1 }{ 5-1 } =\frac { z+3 }{ -8+3 } \)
\(\Rightarrow \frac { x-2 }{ -3 } =\frac { y-1 }{ 4 } =\frac { z+3 }{ -5 } \)
Hence the required plane is parallel to the vector
\(\vec { c } =-3\hat { i } +4\hat { j } -5\hat { k } \)
The parametric form of vector equation of the plane passing through two points \(\vec { a } ,\vec { b } \) parallel to a vector\(\vec { c }\ is\ \vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t\vec { c } ,s,t\in R,\)
\(\vec { r } .2\hat { i } +2\hat { j } +\hat { k } +s(-\hat { i } -4\hat { j } +2\hat { k } )+t(3\hat { i } -4\hat { j } +5\hat { k } )\) s, t ∈ R 
Cartesian form of the plane passing through two points and parallel to a vector is
\(\left| \begin{matrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| =0\)
[∵(x₁,y₁, z₁) is (2, 2, 1), (x₂, y₂, z₂) is (-1, -2, 3) & (c₁, c₂, c₃) is (-3, 4 -5)
\(\Rightarrow \left| \begin{matrix} x-2 & y-2 & z-1 \\ -1 & -4 & 2 \\ -3 & 4 & -5 \end{matrix} \right| =0\)
⇒ (x - 2)(20 - 8) - (y- 2)(5+ 6) + (z-1)(-4 -12) = 0
⇒ (x - 2)(12) - (y - 2)(11) + (z - 1)(-16) = 0
⇒ 12x- 24 -11y + 22 -16z + 16 = 0
⇒ 12x-11y-16z+14 = 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
\(\vec { r } .(\hat { i } -2\hat { j } +4\hat { k } )=20\)
\(Let\quad \vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\Rightarrow (x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } -2\hat { j } +4\hat { k } )=20\)
\(\Rightarrow x=2y+4z=20\)
\(\Rightarrow x-2y+4z-20=0\)

The required plane is parallel to the given line and so it is parallel to the vector \(\vec { c } =\hat { i } +\hat { j } -\hat { k } \) and the plane passes through the points \(\vec { a } =-\hat { i } +2\hat { j } ,\vec { b } =2\hat { i } +2\hat { j } -\hat { k } \)
(i) vector equation of the plane in parametric form is \(\vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t\vec { c } \), where s, t ∈ R
which implies that \(\vec { r } =(-\hat { i } +2\hat { j } )+s(3\hat { i } -\hat { k } )+t(\hat { i } +\hat { j } -\hat { k } )\), where s, t ∈ R
(ii) vector equation of the plane in non-parametric form is \((\vec { r } -\vec { a } ).(\vec { b } -\vec { a } )\times \vec { c } )\) = 0
Now, \((\vec { b } -\vec { a } )\times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 0 & -1 \\ 1 & -1 & -1 \end{matrix} \right| =\hat { i } +2\hat { j } +3\hat { k } \)
we have \((\vec { r } -(-\hat { i } +2\hat { j } ).(\hat { i } +2\hat { j } +3\hat { k } )\) = 0 ⇒ \(\vec { r } .(\hat { i } +2\hat { j } +3\hat { k } )\) = 3
If  \(\vec { r } .(\hat { i } +2\hat { j } +3\hat { k } )\) 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 x + 2y + 3z = 3