Explanation

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