Find the non-parametric vector equation, parametric vector equation and cartesian equation of the plane through the point $\left(-1,3,2\right)$ and perpendicular to the planes $x + 2y + 2z = 5$ and $3x + y + 2z = 8$
Non-parametric vector equation of plane
The equation of a plane passing through a given point $\overrightarrow{a}$ and perpendicular to two given planes $\overrightarrow{r} \cdot \overrightarrow{n_1} = d_1$ and $\overrightarrow{r} \cdot \overrightarrow{n_2} = d_2$ is
$\left(\overrightarrow{r} - \overrightarrow{a}\right) \cdot \left(\overrightarrow{n_1} \times \overrightarrow{n_2}\right) = 0$
The required plane pass through the point $A \left(x_1,y_1,z_1\right) = \left(-1,3,2\right)$
$\therefore$ $\;$ position vector of point $A$ is $\overrightarrow{a} = - \hat{i} + 3 \hat{j} + 2 \hat{k}$
Cartesian equations of given planes are
$x + 2y + 2z = 5$ $\;\;\; \cdots \; (1a)$
$3x + y + 2z = 8$ $\;\;\; \cdots \; (2a)$
$\therefore$ $\;$ Vector equations of the given planes are
$\overrightarrow{r} \cdot \left(\hat{i} + 2 \hat{j} + 2 \hat{k}\right) = 5$ $\;\;\; \cdots \; (1b)$
$\overrightarrow{r} \cdot \left(3 \hat{i} + \hat{j} + 2 \hat{k}\right) = 8$ $\;\;\; \cdots \; (2b)$
Let $\;\;$ $\overrightarrow{n_1} = \hat{i} + 2 \hat{j} + 2 \hat{k}$; $\;\;$ $\overrightarrow{n_2} = 3 \hat{i} + \hat{j} + 2 \hat{k}$; $\;\;$ $d_1 = 5$; $\;\;$ $d_2 = 8$
Now,
$\begin{aligned}
\overrightarrow{n_1} \times \overrightarrow{n_2} & = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 2 \\
3 & 1 & 2
\end{vmatrix} \\\\
& = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}
\end{aligned}$
$\therefore$ $\;$ The non-parametric vector equation of the required plane is
$\left[\overrightarrow{r} - \left(- \hat{i} + 3 \hat{j} + 2 \hat{k}\right)\right] \cdot \left(2 \hat{i} + 4 \hat{j} - 5 \hat{k}\right) = 0$
i.e. $\;$ $\overrightarrow{r} \cdot \left(2 \hat{i} + 4 \hat{j} - 5 \hat{k}\right) = \left(- \hat{i} + 3 \hat{j} + 2 \hat{k}\right) \cdot \left(2 \hat{i} + 4 \hat{j} - 5 \hat{k}\right)$
i.e. $\;$ $\overrightarrow{r} \cdot \left(2 \hat{i} + 4 \hat{j} - 5 \hat{k}\right) = 0$
Parametric vector equation of plane
The normal vector to plane $(1a)$ is $\;\;$ $\overrightarrow{u} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ $\;\;\; \cdots \; (1c)$
The normal vector to plane $(1a)$ is $\;\;$ $\overrightarrow{v} = 3 \hat{i} + \hat{j} + 2 \hat{k}$ $\;\;\; \cdots \; (2c)$
$\therefore$ $\;$ The required plane passes through the point $A$ and is parallel to the vectors $(1c)$ and $(2c)$.
$\therefore$ $\;$ Parametric vector equation of the plane is
$\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{u} + \mu \overrightarrow{v}$ $\;\;$ [$\lambda$ and $\mu$ are scalars]
i.e. $\;$ $\overrightarrow{r} = \left(- \hat{i} + 3 \hat{j} + 2 \hat{k}\right) + \lambda \left(\hat{i} + 2 \hat{j} + 2 \hat{k}\right) + \mu \left(3 \hat{i} + \hat{j} + 2 \hat{k}\right)$
Cartesian equation of plane
Equations of given planes are $(1a)$ and $(2a)$.
They are of the form $\;\;$ $a_1 x + b_y + c_1z + d_1 = 0$ $\;$ and $\;$ $a_2 x + b_2 y + c_2 z + d_2 = 0$ respectively.
$\therefore$ $\;$ $a_1 = 1, \; b_1 = 2, \; c_1 = 2, \; d_1 = -5$ $\;$ and $\;$ $a_2 = 3, \; b_2 = 1, \; c_2 = 2, \; d_2 = - 8$
Equation of the required plane through point $A$ is
$a \left(x + 1\right) + b \left(y - 3\right) + c \left(z - 2\right) = 0$ $\;\;\; \cdots \; (3a)$
$\because$ $\;$ The required plane is perpendicular to the given planes, we have
$aa_1 + bb_1 + c c_1 = 0$
i.e. $\;$ $a + 2b + 2c = 0$ $\;\;\; \cdots \; (3b)$
and $\;$ $a a_2 + b b_2 + c c_2 = 0$
i.e. $\;$ $3a + b + 2c = 0$ $\;\;\; \cdots \; (3c)$
Eliminating $a$, $b$ and $c$ from equations $(3a)$, $(3b)$ and $(3c)$, the required equation of plane is
$\begin{vmatrix}
x + 1 & y - 3 & z - 2 \\
1 & 2 & 2 \\
3 & 1 & 2
\end{vmatrix} = 0$
i.e. $\;$ $2 \left(x + 1\right) + 4 \left(y - 3\right) - 5 \left(z - 2\right) = 0$
i.e. $\;$ $2x + 4y - 5z = 0$