A non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\vec{i}, \vec{i} + \vec{j} $ and the plane determined by the vectors determined by $\vec{i} - \vec{j}, \vec{i} + \vec{k}$. The angle between $\vec{a}$ and $\hat{i}-2\hat{j} + 2\hat{k},$ is

$\frac{\pi}{4 }$

The line of intersection of the two planes is perpendicular to their normals. So, it is parallel to the vector

$\vec{a}= \begin{Bmatrix}\hat{i} × (\hat{i} +\hat{j})\end{Bmatrix}×\begin{Bmatrix}(\hat{i} - \hat{j})×(\hat{i} +\hat{k})\end{Bmatrix}$

$⇒ \vec{a}= \hat{k} × (-\hat{j} +\hat{k} - \hat{i})=\hat{i} +\hat{j}$

Let θ be the angle between $\vec{a}$ and $\vec{b}= \hat{i} - 2\hat{j}+2\hat{k}$. Then, 

$cos \theta = \frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}=\frac{3}{\sqrt{2}× 3}=\frac{1}{\sqrt{2}}⇒ \theta = \frac{\pi}{4}$