Explanation

(c)
let $\overrightarrow{\mathbf{a}} = l\widehat{\dot{\mathbf{i}}} + m\widehat{\dot{\mathbf{j}}} + n\widehat{\mathbf{k}}\ makes\ an\ angle\frac{\pi}{4}with\ z$-axis
Also, $l^{2} + m^{2} + n^{2} = 1$
$Here,\ n = \cos{\frac{\pi}{4} = \frac{1}{\sqrt{2}},\ \ l^{2} + m^{2} = \frac{1}{2}}\ \ \ \ \ \ \ldots\ldots(i)$
$\therefore\ \ \overrightarrow{\mathbf{a}} = l\widehat{\dot{\mathbf{i}}} + m\widehat{\dot{\mathbf{j}}} + \frac{\widehat{\mathbf{k}}}{\sqrt{2}}$
$\Longrightarrow \ \overrightarrow{\mathbf{a}} + \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} = (l + 1)\widehat{\dot{\mathbf{i}}}(m + 1)\widehat{\dot{\mathbf{j}}} + \frac{\widehat{\mathbf{k}}}{\sqrt{2}}$
$\Longrightarrow \left| \overrightarrow{\mathbf{a}} + \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} \right| = \sqrt{(l + 1)^{2} + (m + 1)^{2} + \left( \frac{1}{\sqrt{2}} \right)^{2}}$
$\Longrightarrow 1 = l^{2} + m^{2} + 2 + 2l + 2m + \frac{1}{2}$
$\Longrightarrow l + m = - 1\ \$(From Eq. (i)
$\Longrightarrow \ l^{2} + m^{2} + 2lm = 1$
$\Longrightarrow 2lm = \frac{1}{2}$
$\Longrightarrow l = m = - \frac{1}{2}$
$\left( \because l = m = \frac{1}{2},\ is\ not\ satisfied\ the\ given\ equition \right)$
$\therefore\ \overrightarrow{\mathbf{a}} = - \frac{\widehat{\dot{\mathbf{i}}}}{2} - \frac{\widehat{\dot{\mathbf{j}}}}{2} + \frac{\widehat{\mathbf{k}}}{\sqrt{2}}$