Find the vector and cartesian equation of the plane through the point $\left(1,3,2\right)$ and parallel to the lines $\dfrac{x + 1}{2} = \dfrac{y + 2}{-1} = \dfrac{z+3}{3}$ and $\dfrac{x - 2}{1} = \dfrac{y + 1}{2} = \dfrac{z + 2}{2}$
The required plane passes through the point $A \left(1,3,2\right)$
$\therefore$ $\;$ position vector of point $A$ is $\overrightarrow{a} = \hat{i} + 3 \hat{j} + 2 \hat{k}$
The given equations of lines are
$\dfrac{x + 1}{2} = \dfrac{y + 2}{-1} = \dfrac{z+3}{3}$ $\;\;\; \cdots \; (1a)$
$\dfrac{x - 2}{1} = \dfrac{y + 1}{2} = \dfrac{z + 2}{2}$ $\;\;\; \cdots \; (2a)$
Vector form of equation $(1a)$ is
$\overrightarrow{r} = \left(-\hat{i}-2 \hat{j} -3 \hat{k}\right) + \lambda \left(2 \hat{i} - \hat{j} + 3 \hat{k}\right)$ $\;\;\; \cdots \; (1b)$
Vector form of equation $(2a)$ is
$\overrightarrow{r} = \left(2 \hat{i} - \hat{j} - 2 \hat{k}\right) + \mu \left(\hat{i} + 2 \hat{j} + 2 \hat{k}\right)$ $\;\;\; \cdots \; (2b)$
Required vector equation of plane in vector form is
$\overrightarrow{r} = \overrightarrow{a} + s \overrightarrow{u} + t \overrightarrow{v}$
where $\;$ $\overrightarrow{u} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ $\;$ and $\;$ $\overrightarrow{v} = \hat{i} + 2 \hat{j} + 2 \hat{k}$; $\;\;\;$ $s$ and $t$ are scalars.
i.e. $\;$ $\overrightarrow{r} = \hat{i} + 3 \hat{j} + 2 \hat{k} + s \left(2 \hat{i} - \hat{j} + 3 \hat{k}\right) + t \left(\hat{i} + 2 \hat{j} + 2 \hat{k}\right)$
Now,
$A \left(x_1, y_1, z_1\right) = \left(1, 3, 2\right)$
direction ratios of line $(1a)$ are $\; \ell_1 = 2, \; m_1 = -1, \; n_1 = 3$
direction ratios of line $(2a)$ are $\; \ell_2 = 1, \; m_2 = 2, \; n_2 = 2$
Equation of the required plane in cartesian coordinates is
$\begin{vmatrix}
x - x_1 & y - y_1 & z - z_1 \\
\ell_1 & m_1 & n_1 \\
\ell_2 & m_2 & n_2
\end{vmatrix} = 0$
i.e. $\;$ $\begin{vmatrix}
x - 1 & y - 3 & z - 2 \\
2 & -1 & 3 \\
1 & 2 & 2
\end{vmatrix} = 0$
i.e. $\;$ $\left(x - 1\right) \left(-2 - 6\right) - \left(y - 3\right) \left(4 - 3\right) + \left(z - 2\right) \left(4 + 1\right) = 0$
i.e. $\;$ $-8x + 8 - y + 3 + 5z - 10 =0$
i.e. $\;$ $8x + y - 5z = 1$