Show that the lines $\;$ $\overrightarrow{r} = \left(3 \hat{i} + 5 \hat{j} + 7 \hat{k}\right) + t \left(\hat{i} - 2 \hat{j} + \hat{k}\right)$ $\;$ and $\;$ $\overrightarrow{r} = \left(\hat{i} + \hat{j} + \hat{k}\right) + s \left(7 \hat{i} + 6 \hat{j} + 7 \hat{k}\right)$ $\;$ are skew lines and find the shortest distance between the lines.
The given lines are
$\overrightarrow{r} = \left(3 \hat{i} + 5 \hat{j} + 7 \hat{k}\right) + t \left(\hat{i} - 2 \hat{j} + \hat{k}\right)$ $\;\;\; \cdots \; (1a)$
$\overrightarrow{r} = \left(\hat{i} + \hat{j} + \hat{k}\right) + s \left(7 \hat{i} + 6 \hat{j} + 7 \hat{k}\right)$ $\;\;\; \cdots \; (1b)$
Comparing equations $(1a)$ and $(1b)$ with
$\overrightarrow{r} = \overrightarrow{a_1} + t \; \overrightarrow{u}$ $\;$ and $\;$ $\overrightarrow{r} = \overrightarrow{a_2} + s \; \overrightarrow{v}$ $\;$ respectively, we have
$\overrightarrow{a_1} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k}$ $\;\;\; \cdots \; (2a)$
$\overrightarrow{a_2} = \hat{i} + \hat{j} + \hat{k}$ $\;\;\; \cdots \; (2b)$
$\overrightarrow{u} = \hat{i} - 2 \hat{j} + \hat{k}$ $\;\;\; \cdots \; (2c)$
$\overrightarrow{v} = 7 \hat{i} + 6 \hat{j} + 7 \hat{k}$ $\;\;\; \cdots \; (2d)$
Condition for two lines are skew lines to be skew: $\;$ $\left[\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \; \overrightarrow{u} \; \overrightarrow{v}\right] \neq 0$
Now,
$\begin{aligned}
\overrightarrow{a_2} - \overrightarrow{a_1} & = \left(\hat{i} + \hat{j} + \hat{k}\right) - \left(3 \hat{i} + 5 \hat{j} + 7 \hat{k}\right) \\\\
& = - 2 \hat{i} - 4 \hat{j} - 6 \hat{k}
\end{aligned}$
$\begin{aligned}
\therefore \; \left[\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \; \overrightarrow{u} \; \overrightarrow{v}\right] & = \begin{vmatrix}
-2 & -4 & -6 \\
1 & -2 & 1 \\
7 & 6 & 7
\end{vmatrix} \\\\
& = -2 \left(-14 - 6\right) + 4 \left(7 - 7\right) - 6 \left(6 + 14\right) \\\\
& = 40 - 120 \\\\
& = - 80 \neq 0
\end{aligned}$
$\because$ $\;$ $\left[\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \; \overrightarrow{u} \; \overrightarrow{v}\right] \neq 0$, the given lines are skew lines.
Shortest distance between the lines $= d = \dfrac{\left|\left[\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \; \overrightarrow{u} \; \overrightarrow{v}\right]\right|}{\left|\overrightarrow{u} \times \overrightarrow{v}\right|}$ $\;\;\; \cdots \; (3)$
Now, $\left|\left[\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \; \overrightarrow{u} \; \overrightarrow{v}\right]\right| = \left|-80\right| = 80$ $\;\;\; \cdots \; (4a)$
$\begin{aligned}
\overrightarrow{u} \times \overrightarrow{v} & = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 1 \\
7 & 6 & 7
\end{vmatrix} \\\\
& = \hat{i} \left(-14 - 6\right) - \hat{j} \left(7 - 7\right) + \hat{k} \left(6 + 14\right) \\\\
& = - 20 \hat{i} + 20 \hat{k}
\end{aligned}$
$\therefore$ $\;$ $\left|\overrightarrow{u} \times \overrightarrow{v}\right| = \sqrt{\left(-20\right)^2 + \left(20\right)^2} = \sqrt{800} = 20 \sqrt{2}$ $\;\;\; \cdots \; (4b)$
$\therefore$ $\;$ From equations $(3)$, $(4a)$ and $(4b)$, the shortest distance between the lines is
$d = \dfrac{80}{20 \sqrt{2}} = 2 \sqrt{2}$