Find the cartesian equation of the plane which contains the lines $\dfrac{x+1}{2} = \dfrac{y - 2}{-3} = \dfrac{z - 3}{4}$ and $\dfrac{x - 4}{3} = \dfrac{y - 1}{2} = z - 8$
Equations of the given lines are
$L_1: \;\; \dfrac{x+1}{2} = \dfrac{y - 2}{-3} = \dfrac{z - 3}{4}$
$L_2: \;\; \dfrac{x - 4}{3} = \dfrac{y - 1}{2} = \dfrac{z - 8}{1}$
Lines $L_1$ and $L_2$ are of the form
$\dfrac{x - x_1}{a_1} = \dfrac{y - y_1}{b_1} = \dfrac{z - z_1}{c_1}$ $\;$ and $\;$ $\dfrac{x - x_2}{a_2} = \dfrac{y - y_2}{b_2} = \dfrac{z - z_2}{c_2}$ $\;$ respectively.
Here
$x_1 = -1, \; y_1 = 2, \; z_1 = 3$; $\;$ $x_2 = 4, \; y_2 = 1, \; z_2 = 8$
$a_1 = 2, \; b_1 = -3, \; c_1 = 4$; $\;$ $a_2 = 3, \; b_2 = 2, \; c_2 = 1$
Now,
$\begin{aligned}
\begin{vmatrix}
x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{vmatrix} & = \begin{vmatrix}
5 & -1 & 5 \\
2 & -3 & 4 \\
3 & 2 & 1
\end{vmatrix} \\\\
& = 5 \left(-3 - 8\right) + \left(2 - 12\right) + 5 \left(4 + 9\right) \\\\
& = -55 - 10 + 65 = 0
\end{aligned}$
$\therefore$ $\;$ The given lines are coplanar.
$\therefore$ $\;$ The equation of plane containing the given lines is
$\begin{vmatrix}
x - x_1 & y - y_1 & z - z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{vmatrix} = 0$
i.e. $\;$ $\begin{vmatrix}
x +1 & y - 2 & z - 3 \\
2 & -3 & 4 \\
3 & 2 & 1
\end{vmatrix} = 0$
i.e. $\;$ $\left(x + 1\right) \left(-3 - 8\right) - \left(y - 2\right) \left(2 - 12\right) + \left(z - 3\right) \left(4 + 9\right) = 0$
i.e. $\;$ $- 11x + 10 y + 13 z - 70 = 0$
i.e. $\;$ $11x - 10 y - 13 z + 70 = 0$