Explanation

(d)
A vector normal to first plane is ${\overrightarrow{\mathbf{n}}}_{1} = \widehat{\dot{\mathbf{i}}} \times \left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} \right) = \widehat{\mathbf{k}}$
A vector normal to second plane is ${\overrightarrow{\mathbf{n}}}_{2}$
$= \left( \widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} \right) \times \left( \widehat{\dot{\mathbf{i}}} + \widehat{\mathbf{k}} \right) = - \ \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} - \widehat{\dot{\mathbf{i}}}$
Since, $\overrightarrow{\mathbf{a}}$ will be parallel to ${\overrightarrow{\mathbf{n}}}_{1} \times {\overrightarrow{\mathbf{n}}}_{2} = \widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}}$
Let θ be the angle between $\overrightarrow{\mathbf{a}}\$ and $\widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}$
$\therefore\cos\theta\frac{\left( \widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} \right) \bullet \left( \widehat{\dot{\mathbf{i}}} - 2\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right)}{\sqrt{1^{2} + 1^{2}}\ \sqrt{1^{2} + 2^{2} + 2^{2}}}$
$= \frac{1 + 2}{\sqrt{2} \bullet 3} = \frac{1}{\sqrt{2}}$
$\Longrightarrow \ \theta = \frac{\pi}{4}$