Explanation

(a)
$\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
1 & 1 & 1 \\
1 & 1 & 0
\end{matrix} \right|$
$= - \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}}$
$\Longrightarrow \left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \times \overrightarrow{\mathbf{c}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
- 1 & 1 & 0 \\
1 & 0 & 0
\end{matrix} \right| = - \widehat{\mathbf{k}}$
Now, $\lambda\ \overrightarrow{\mathbf{a}} + \mu\ \overrightarrow{\mathbf{b}} = \lambda\left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) + \mu\left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} \right)$
$= (\lambda\ + \mu)\widehat{\dot{\mathbf{i}}} + (\lambda + \mu)\widehat{\dot{\mathbf{j}}} + \lambda\widehat{\mathbf{k}}$
$\because\ \lambda\ \overrightarrow{\mathbf{a}} + \mu\ \overrightarrow{\mathbf{b}} = \left( \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right) \times \overrightarrow{\mathbf{c}}$
$\Longrightarrow (\lambda + \mu)\widehat{\dot{\mathbf{i}}} + (\lambda + \mu)\widehat{\dot{\mathbf{j}}} + \lambda\widehat{\mathbf{k}} = - \widehat{\mathbf{k}}$
On equating the coefficient of $\widehat{\dot{\mathbf{i}}}\$ we get $\lambda + \mu = 0$