Explanation

(b)
Let $\overrightarrow{\mathbf{d}}$=$d_{1}\widehat{\dot{\mathbf{i}}} + d_{2}\widehat{\dot{\mathbf{j}}} + d_{3}\widehat{\mathbf{k}}$
$\overrightarrow{\mathbf{a}} \bullet \ \overrightarrow{\mathbf{d}} = d_{1} - d_{2} = 0\ \Longrightarrow \ d_{1} = d_{2}$ ….(i)
Also, $\overrightarrow{\mathbf{d}}$ is a unit vector.
$\Longrightarrow \ d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = 1$ ….(ii)
$Also,\ \left\lbrack \overrightarrow{\mathbf{b}}\ \overrightarrow{\mathbf{c}}\ \overrightarrow{\mathbf{d}} \right\rbrack = 0 \Longrightarrow \left| \begin{matrix}
0 & 1 & - 1 \\
- 1 & 0 & 1 \\
d_{1} & d_{2} & d_{3}
\end{matrix} \right| = 0$
$\Longrightarrow \ - 1\left( - d_{3} - d_{1} \right) - 1\left( - d_{2} \right) = 0$
$\Longrightarrow \ d_{1} + d_{2} + d_{3} = 0 \Longrightarrow 2d_{1} + d_{3} = 0$ [from Eq. (i) ]
$\Longrightarrow \ d_{3} = - 2d_{1}$ ….(iii)
Using Eqs. (iii) and (i) in Eq. (ii), we get
$d_{1}^{2} + d_{1}^{2} + 4d_{1}^{2} = 1\ \ \Longrightarrow$ $d_{1} = \pm \frac{1}{\sqrt{6}}$
$\therefore\ \ d_{2} = \pm \frac{1}{\sqrt{6}}$
$and{\ \ d}_{3} = \mp \frac{2}{\sqrt{6}}\$
Hence, required vector is
$\pm \frac{1}{\sqrt{6}}(\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}})$