Explanation

(a)
$\overrightarrow{AB} = \overrightarrow{\beta} - \overrightarrow{\alpha} = - 2\widehat{i} - 3\widehat{j} - 6\widehat{k}$
Equation of the plane passing through $B$ and perpendicular to $AB$ is
$\left( \overrightarrow{r} - \overrightarrow{OB} \right) \bullet \overrightarrow{AB} = 0$
$\overrightarrow{r}.(2\widehat{i} + 3\widehat{j} + 6\widehat{k}) + 28 = 0$
Hence the required distance from $\overrightarrow{r} = - \widehat{i} + \widehat{j} + \widehat{k}$
$= \left| \frac{\left( - \widehat{i} + \widehat{j} + \widehat{k} \right).\left( 2\widehat{i} + 3\widehat{j} + 6\widehat{k} \right) + 28}{|2\widehat{i} + 3\widehat{j} + 6\widehat{k}|} \right| = \left| \frac{- 2 + 3 + 6 + 28}{7} \right| = 5\ units$