Explanation

(d)
Let $\overrightarrow{\mathbf{c}} = \overrightarrow{\mathbf{a}} + \lambda + \overrightarrow{\mathbf{b}} = (1 + \lambda)\widehat{\dot{\mathbf{i}}} + (1 - \lambda)\widehat{\dot{\mathbf{j}}} + (\lambda - 1)\widehat{\mathbf{k}}$
Also, $\overrightarrow{\mathbf{c}} \bullet \overrightarrow{\mathbf{a}} = 0$
$\Longrightarrow \left\lbrack (1 + \lambda)\widehat{\dot{\mathbf{i}}} + (1 - \lambda)\widehat{\dot{\mathbf{j}}} + (\lambda - 1)\widehat{\mathbf{k}} \right\rbrack \bullet \left\lbrack \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}} \right\rbrack = 0$
$\Longrightarrow 1 + \lambda + 1 - \lambda - \lambda + 1 = 0$
$\Longrightarrow \ \lambda = 3$
$\therefore\ \ \overrightarrow{\mathbf{c}} = 4\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}\$
$\Longrightarrow \ \overrightarrow{\mathbf{c}} = \pm \frac{2\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}}{\sqrt{6}}$