Explanation

(d)
Given lines are $\overrightarrow{r} = 3\widehat{i} + 8\widehat{j} + 3\widehat{k} + l(3\widehat{i} - \widehat{j} + \widehat{k})$ and $\overrightarrow{r} = - 3\widehat{i} - 7\widehat{j} + 6\widehat{k} + m( - 3\widehat{i} + 2\widehat{j} + 4\widehat{k})$
Required shortest distance
$= \frac{\begin{array}{r}
|\left( 6\widehat{i} + 15\widehat{j} - 3\widehat{k} \right).((3\widehat{i} - \widehat{j} + \widehat{k}) \times \\
( - 3\widehat{i} + 2\widehat{j} + 4\widehat{k}))|
\end{array}}{|(3\widehat{i} - \widehat{j} + \widehat{k}) \times ( - 3\widehat{i} + 2\widehat{j} + 4\widehat{k})|}$
$= \frac{|\left( 6\widehat{i} + 15\widehat{j} - 3\widehat{k} \right).\left( - 6\widehat{i} - 15\widehat{j} + 3\widehat{k} \right)|}{|( - 6\widehat{i} - 15\widehat{j} + 3\widehat{k})|}$
$= \frac{36 + 225 + 9}{\sqrt{36 + 225 + 9}} = \frac{270}{\sqrt{270}} = \sqrt{270} = 3\sqrt{30}$