Explanation

(d)
$\overrightarrow{\mathbf{AC}} = \left( a\widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) - \left( 2\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) = (a - 2)\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}}$
and $\overrightarrow{\mathbf{BC}} = \left( a\widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) - \left( \widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} - 5\widehat{\mathbf{k}} \right) = (a - 1)\widehat{\dot{\mathbf{i}}} + 6\widehat{\mathbf{k}}$
Since, the $\mathrm{\Delta}ABC$ is right angled at $C,$ then
$\overrightarrow{\mathbf{AC}} \bullet \overrightarrow{\mathbf{BC}} = 0$
$\Longrightarrow \ \left\{ (a - 2)\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} \right\}\ \bullet \left\{ (a - 1)\widehat{\dot{\mathbf{i}}} + 6\widehat{\mathbf{k}} \right\} = 0$
$\Longrightarrow (a - 2)(a - 1) = 0 \Longrightarrow a = 1and\ 2$