Explanation

(c)
A unit perpendicular to the plane $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}} = \frac{\overrightarrow{\mathbf{a}}\mathbf{\times}\overrightarrow{\mathbf{b}}}{\left| \ \overrightarrow{\mathbf{a}}\mathbf{\times}\overrightarrow{\mathbf{b}}\ \right|}$
Now, $\overrightarrow{\mathbf{a}}\mathbf{\times}\overrightarrow{\mathbf{b}} = \left| \begin{matrix}
\widehat{\mathbf{i}} & \widehat{\mathbf{j}} & \widehat{\mathbf{k}} \\
2 & - 6 & - 3 \\
4 & 3 & - 1
\end{matrix} \right|$
$= \widehat{\mathbf{i}}(6 + 9) - \widehat{\mathbf{j}}( - 2 + 12) + \widehat{\mathbf{k}}(6 + 24)$
$= 15\widehat{\mathbf{i}} - 10\widehat{\mathbf{j}} + 30\widehat{\mathbf{k}}$
and $\left| \overrightarrow{\mathbf{a}}\mathbf{\times}\overrightarrow{\mathbf{b}} \right| = \sqrt{15^{2} + ( - 10)^{2} + (30)^{2}}$
$= \sqrt{1225} = 35$
$\therefore$ Required vector $= \frac{15\widehat{\mathbf{i}} - 10\widehat{\mathbf{j}} + 30\widehat{\mathbf{k}}}{35} = \frac{3\widehat{\mathbf{i}} - 2\widehat{\mathbf{j}} + 6\widehat{\mathbf{k}}}{7}$