Find the shortest distance between the parallel lines $\;$ $\dfrac{x - 1}{-1} = \dfrac{y}{3} = \dfrac{z + 3}{2}$ $\;$ and $\;$ $\dfrac{x - 3}{-1} = \dfrac{y + 1}{3} = \dfrac{z - 1}{2}$
The equations of the lines in Cartesian form are
$\dfrac{x - 1}{-1} = \dfrac{y}{3} = \dfrac{z + 3}{2}$ $\;\;\; \cdots \; (1a)$
$\dfrac{x - 3}{-1} = \dfrac{y + 1}{3} = \dfrac{z - 1}{2}$ $\;\;\; \cdots \; (1b)$
Vector form of equation $(1a)$ is
$\overrightarrow{r} = \left(\hat{i} - 3 \hat{k}\right) + \lambda \left(-\hat{i} + 3 \hat{j} + 2 \hat{k}\right)$ $\;\;\; \cdots \; (2a)$
Vector form of equation $(1b)$ is
$\overrightarrow{r} = \left(3 \hat{i} - \hat{j} + \hat{k} \right) + \mu \left(- \hat{i} + 3 \hat{j} + 2 \hat{k}\right)$ $\;\;\; \cdots \; (2b)$
Comparing equations $(2a)$ and $(2b)$ with
$\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{u}$ $\;$ and $\;$ $\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{u}$ $\;$ respectively, we have
$\overrightarrow{a_1} = \hat{i} - 3 \hat{k}$ $\;\;\; \cdots \; (3a)$
$\overrightarrow{a_2} = 3 \hat{i} - \hat{j} + \hat{k}$ $\;\;\; \cdots \; (3b)$
$\overrightarrow{u} = - \hat{i} + 3 \hat{j} + 2 \hat{k}$ $\;\;\; \cdots \; (3c)$
Shortest distance between parallel lines $= d = \dfrac{\left|\overrightarrow{u} \times \left(\overrightarrow{a_2} - \overrightarrow{a_1}\right)\right|}{\left|\overrightarrow{u}\right|}$ $\;\;\; \cdots \; (4)$
$\begin{aligned}
\overrightarrow{a_2} - \overrightarrow{a_1} & = \left(3 \hat{i} - \hat{j} + \hat{k}\right) - \left(\hat{i} - 3 \hat{k}\right) \\\\
& = 2 \hat{i} - \hat{j} + 4 \hat{k}
\end{aligned}$
$\begin{aligned}
\therefore \; \overrightarrow{u} \times \left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) & = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 3 & 2 \\
2 & -1 & 4
\end{vmatrix} \\\\
& = \hat{i} \left(12 + 2\right) - \hat{j} \left(-4 - 4\right) + \hat{k} \left(1 - 6\right) \\\\
& = 14 \hat{i} + 8 \hat{j} - 5 \hat{k}
\end{aligned}$
$\therefore$ $\;$ $\left|\overrightarrow{u} \times \left(\overrightarrow{a_2} - \overrightarrow{a_1}\right)\right| = \sqrt{\left(14\right)^2 + \left(8\right)^2 + \left(-5\right)^2} = \sqrt{196 + 64 + 25} = \sqrt{285}$
$\left|\overrightarrow{u}\right| = \sqrt{\left(-1\right)^2 + \left(3\right)^2 + \left(2\right)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$ $\;\;\; \cdots \; (5b)$
Substituting equations $(5a)$ and $(5b)$ in equation $(4)$, we have
$d = \sqrt{\dfrac{285}{14}}$