Find the cartesian equation of the plane containing the line $\dfrac{x - 2}{2} = \dfrac{y - 2}{3} = \dfrac{z - 1}{3}$ and parallel to the line $\dfrac{x + 1}{3} = \dfrac{y - 1}{2} = \dfrac{z + 1}{1}$
The equation of any plane through the line $\;$ $\dfrac{x - 2}{2} = \dfrac{y - 2}{3} = \dfrac{z -1}{3}$ $\;$ is
$A \left(x - 2\right) + B \left(y - 2\right) + C \left(z - 1\right) = 0$ $\;\;\; \cdots \; (1)$
where $\;$ $2 A + 3 B + 3 C = 0$ $\;\;\; \cdots \; (2)$
$\because$ $\;$ Plane given by equation $(1)$ is parallel to the line $\;$ $\dfrac{x + 1}{3} = \dfrac{y - 1}{2} = \dfrac{z + 1}{1}$,
the normal to equation $(1)$ is perpendicular to $(1)$ and has direction ratios $3, \; 2, \; 1$.
$\therefore$ $\;$ $3 A + 2 B + C = 0$ $\;\;\; \cdots \; (3)$
Solving equations $(2)$ and $(3)$ we get
$\dfrac{A}{\begin{vmatrix}
3 & 3 \\
2 & 1
\end{vmatrix}} = \dfrac{-B}{\begin{vmatrix}
2 & 3 \\
3 & 1
\end{vmatrix}} = \dfrac{C}{\begin{vmatrix}
2 & 3 \\
3 & 2
\end{vmatrix}} = k$ (say)
i.e. $\dfrac{A}{3 - 6} = \dfrac{B}{9 - 2} = \dfrac{C}{4 - 9} = k$
i.e. $A = - 3k$, $\;$ $B = 7 k$, $\;$ $C = -5k$
Substituting the values of $A$, $B$ and $C$ in equation $(1)$, we have
$-3k \left(x - 2\right) + 7 k \left(y - 2\right) - 5 k \left(z - 1\right) = 0$
$\implies$ $-3x + 7y - 5z = 3$
i.e. $\;$ $3x - 7y + 5z + 3 = 0$ $\;\;\; \cdots \; (4)$
Equation $(4)$ is the required equation of the plane in cartesian form.