Explanation

(c)
Now, $\left( \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}} \right) \times \left( \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}} \right) = \left| \begin{matrix}
\widehat{\mathbf{i}} & \widehat{\mathbf{j}} & \widehat{\mathbf{k}} \\
1 & 1 & 1 \\
1 & 1 & 0
\end{matrix} \right|$
$= \widehat{\mathbf{i}}( - 1) + \widehat{\mathbf{j}}(1) + \widehat{\mathbf{k}}(0) = - \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}$
and |$\left( \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}} \right) \times \left( \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}} \right)| = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
Vector perpendicular to both of the vectors $\widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}}$ and $\widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}$
$= \frac{\left( \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}} \right)\mathbf{\times}\left( \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}} \right)}{\left| \left( \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\mathbf{+}\widehat{\mathbf{k}} \right)\mathbf{\times}\left( \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}} \right) \right|}$
$= \frac{- \widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}}{\sqrt{2}} = \frac{- 1}{\sqrt{2}}(\widehat{\mathbf{i}}\mathbf{-}\widehat{\mathbf{j}})$
$= c\left( \widehat{\mathbf{i}}\mathbf{-}\widehat{\mathbf{j}} \right),c$ is a scalar.