Find the parametric vector equation and cartesian equation of the plane through the points $\left(1,2,3\right)$ and $\left(2,3,1\right)$ and perpendicular to the plane $3x - 2y + 4z - 5 = 0$.
Let the given points be $A \left(x_1, y_1, z_1\right) = \left(1,2,3\right)$ and $B \left(x_2, y_2, z_2\right) = \left(2, 3, 1\right)$
position vector of point $A = \overrightarrow{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ $\;\;\; \cdots \; (1a)$
position vector of point $B = \overrightarrow{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ $\;\;\; \cdots \; (1b)$
Equation of given plane is $3x - 2y + 4z - 5 = 0$ $\;\;\; \cdots \; (2a)$
$\because$ $\;$ The given plane is perpendicular to the required plane, the normal vector to the given plane is parallel to the required plane.
Normal vector to the given plane [equation $(2a)$] is
$\overrightarrow{v} = 3 \hat{i} - 2 \hat{j} + 4 \hat{k}$ $\;\;\; \cdots \; (2b)$
Parametric vector equation of a plane passing through two given points [$(1a)$ and $(1b)$] and parallel to the vector $(2b)$ is
$\overrightarrow{r} = \left(1 - s\right) \overrightarrow{a} + s \overrightarrow{b} + t \overrightarrow{v}$ $\;\;$ where $s$ and $t$ are scalars.
$\begin{aligned}
i.e. \;\; \overrightarrow{r} & = \left(1 - s\right) \left(\hat{i} + 2 \hat{j} + 3 \hat{k}\right) + s \left(2 \hat{i} + 3 \hat{j} + \hat{k}\right) + t \left(3 \hat{i} - 2 \hat{j} + 4 \hat{k}\right) \\\\
& = \left(\hat{i} + 2 \hat{j} + 3 \hat{k}\right) + s \left[\left(2 \hat{i} + 3 \hat{j} + \hat{k}\right) - \left(\hat{i} + 2 \hat{j} + 3 \hat{k}\right)\right] + t \left(3 \hat{i} - 2 \hat{j} + 4 \hat{k}\right) \\\\
& = \left(\hat{i} + 2 \hat{j} + 3 \hat{k}\right) + s \left(\hat{i} + \hat{j} - 2 \hat{k}\right) + t \left(3 \hat{i} - 2 \hat{j} + 4 \hat{k}\right) \;\;\; \cdots \; (3)
\end{aligned}$
Equation $(3)$ is the parametric vector equation of the required plane.
For the given plane [equation $(2a)$], $\;$ $\ell_1 = 3, \; m_1 = -2, \; n_1 = 4$
Cartesian equation of the plane is
$\begin{vmatrix}
x - x_1 & y - y_1 & z - z_1 \\
x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\
\ell_1 & m_1 & n_1
\end{vmatrix} = 0$
i.e. $\;$ $\begin{vmatrix}
x - 1 & y - 2 & z - 3 \\
2 - 1 & 3 - 2 & 1 - 3 \\
3 & -2 & 4
\end{vmatrix} = 0$
i.e. $\;$ $\begin{vmatrix}
x - 1 & y - 2 & z - 3 \\
1 & 1 & -2 \\
3 & -2 & 4
\end{vmatrix} = 0$
i.e. $\;$ $\left(x - 1\right) \left(4 - 4\right) - \left(y - 2\right) \left(4 + 6\right) + \left(z - 3\right) \left(-2 -3\right) = 0$
i.e. $\;$ $- 10 \left(y - 2\right) - 5 \left(z - 3\right) = 0$
i.e. $\;$ $2y + z - 7 = 0$ $\;\;\; \cdots \; (4)$
Equation $(4)$ is the cartesian equation of the required plane.