Explanation

(c)
Let ${\overrightarrow{\mathbf{n}}}_{1}\$ and ${\overrightarrow{\mathbf{n}}}_{2}$ be the vectors normal to the plane determined by $\widehat{\dot{\mathbf{i}}},\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}}$ and $\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}},$ $\widehat{\dot{\mathbf{i}}} - \widehat{\mathbf{k}}$ respectively
$\therefore\ {\overrightarrow{\mathbf{n}}}_{1} = \widehat{\dot{\mathbf{i}}} \times \left( \widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} \right) = - \widehat{\mathbf{k}}$
and ${\overrightarrow{\mathbf{n}}}_{2} = \left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} \right) \times \left( \widehat{\dot{\mathbf{i}}} - \widehat{\mathbf{k}} \right) = - \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} - \widehat{\mathbf{k}}$
Since, $\overrightarrow{\mathbf{a}}$ is parallel to the line of intersection of the planes determined by the given planes.
${\therefore\ \ \ \overrightarrow{\mathbf{a}}||({\overrightarrow{\mathbf{n}}}_{1} \times {\overrightarrow{\mathbf{n}}}_{2})
}{\Longrightarrow \overrightarrow{\mathbf{a}} = \lambda\left( {\overrightarrow{\mathbf{n}}}_{1} \times {\overrightarrow{\mathbf{n}}}_{2} \right) = \lambda\left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} \right)}$
Let $\theta$ be the angle between $\overrightarrow{\mathbf{a}}$ and $\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}}$
$\therefore\cos{\theta = \frac{\lambda(\left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} \right) \bullet (\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} - 2\widehat{\mathbf{k}})}{\sqrt{\lambda^{2} + \lambda^{2}}\ \sqrt{1 + 4 + 4}}}$
$= \frac{\lambda(1 + 2)}{\sqrt{2}\lambda \times 3} = \frac{1}{\sqrt{2}}$
$\Longrightarrow \ \theta = \frac{\pi}{4}$