Find the parametric form of vector equation and cartesian equation of the plane passing through the points with position vectors $\;$ $3 \hat{i} + 4 \hat{j} + 2 \hat{k}$, $\;$ $2 \hat{i} - 2 \hat{j} - \hat{k}$ $\;$ and $\;$ $7 \hat{i} + \hat{k}$.
Let the position vectors of the three given points be
$\overrightarrow{a} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k}$, $\;\;\;$ $\overrightarrow{b} = 2 \hat{i} - 2 \hat{j} - \hat{k}$, $\;\;\;$ $\overrightarrow{c} = 7 \hat{i} + \hat{k}$
Let $\;$ $\lambda$ $\;$ and $\;$ $\mu$ $\;$ be two scalars.
Parametric form of vector equation of a plane passing through three non-collinear points is
$\overrightarrow{r} = \left(1 - \lambda - \mu\right) \overrightarrow{a} + \lambda \overrightarrow{b} + \mu \overrightarrow{c}$
$\begin{aligned}
i.e. \; \overrightarrow{r} & = \left(1 - \lambda - \mu\right) \left(3 \hat{i} + 4 \hat{j} + 2 \hat{k}\right) + \lambda \left(2 \hat{i} - 2 \hat{j} - \hat{k}\right) + \mu \left(7 \hat{i} + \hat{k}\right) \\\\
& = \left(3 \hat{i} + 4 \hat{j} + 2 \hat{k}\right) + \lambda \left[\left(2 \hat{i} - 2 \hat{j} - \hat{k}\right) - \left(3 \hat{i} + 4 \hat{j} + 2 \hat{k}\right)\right] \\\\
& \hspace{4cm} + \mu \left[\left(7 \hat{i} + \hat{k}\right) - \left(3 \hat{i} + 4 \hat{j} + 2 \hat{k}\right)\right] \\\\
& = \left(3 \hat{i} + 4 \hat{j} + 2 \hat{k}\right) + \lambda \left(- \hat{i} - 6 \hat{j} - 3 \hat{k}\right) + \mu \left(4 \hat{i} - 4 \hat{j} - \hat{k}\right) \;\;\; \cdots \; (1)
\end{aligned}$
Equation $(1)$ is the parametric form of vector equation of the required plane.
Position vector $\overrightarrow{a}$ represents the point $A \left(x_1, y_1, z_1\right) = \left(3, 4, 2\right)$
Position vector $\overrightarrow{b}$ represents the point $B \left(x_2, y_2, z_2\right) = \left(2, -2, -1\right)$
Position vector $\overrightarrow{c}$ represents the point $C \left(x_3, y_3, z_3\right) = \left(7, 0, 1\right)$
Cartesian equation of the required plane is
$\begin{vmatrix}
x - x_1 & y - y_1 & z - z_1 \\
x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\
x_3 - x_1 & y_3 - y_1 & z_3 - z_1
\end{vmatrix} = 0$
i.e. $\;$ $\begin{vmatrix}
x - 3 & y - 4 & z - 2 \\
2 - 3 & -2 - 4 & -1 - 2 \\
7 - 3 & 0 - 4 & 1 - 2
\end{vmatrix} = 0$
i.e. $\;$ $\begin{vmatrix}
x - 3 & y - 4 & z - 2 \\
-1 & -6 & -3 \\
4 & -4 & -1
\end{vmatrix} = 0$
i.e. $\;$ $\left(x - 3\right) \left(6 - 12\right) - \left(y - 4\right) \left(1 + 12\right) + \left(z - 2\right) \left(4 + 24\right) = 0$
i.e. $\;$ $-6x - 13 y + 28 z + 14 = 0$
i.e. $\;$ $6x + 13 y - 28 z - 14 = 0$ $\;\;\; \cdots \; (2)$
Equation $(2)$ is the cartesian equation of the required plane.