Explanation

(a)
Let unit vector is $a\widehat{\mathbf{i}} + b\widehat{\mathbf{j}} + c\widehat{\mathbf{k}}$.
$\because\ a\widehat{\mathbf{i}} + b\widehat{\mathbf{j}} + c\widehat{\mathbf{k}}$ is perpendicular to $\widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}}$.
Then, $a + b + c = 0$ ...(i)
and $a\widehat{\mathbf{i}} + b\widehat{\mathbf{j}} + c\widehat{\mathbf{k}},\ (\widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}2\widehat{\mathbf{k}})$ and $(\widehat{\mathbf{i}}\mathbf{+}2\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}})$ are coplanar.
$\therefore\ \left| \begin{matrix}
a & b & c \\
1 & 1 & 2 \\
1 & 2 & 1
\end{matrix} \right| = 0$
$\Rightarrow \ - 3a + b + c = 0$ ….(ii)
From Eqs. (i) and (ii), we get
$a = 0$ and $c = - b$
$\because\ a\widehat{\mathbf{i}} + b\widehat{\mathbf{j}} + c\widehat{\mathbf{k}}$ is a unit vector, then
$a^{2} + b^{2} + c^{2} = 1$
$\Rightarrow 0 + b^{2} + b^{2} = 1$
$\Rightarrow b = \frac{1}{\sqrt{2}}$
$\therefore\ a\widehat{\mathbf{i}} + b\widehat{\mathbf{j}} + c\widehat{\mathbf{k}} = \frac{1}{\sqrt{2}}\widehat{\mathbf{j}} - \frac{1}{\sqrt{2}}\widehat{\mathbf{k}} = \frac{\widehat{\mathbf{j}}\mathbf{-}\widehat{\mathbf{k}}}{\sqrt{2}}$