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

$\frac{π}{4}$

Let $\vec{n_1}$ and $\vec{n_2}$ be the vectors normal to the planes determined by $\hat i, \hat i +\hat j$ and $\hat i -\hat j,\hat i+\hat k$ respectively. Then,

$\vec{n_1} =\hat i× (\hat i +\hat j) =\hat k$ and $\vec{n_2} =(\hat i -\hat j)×(\hat i +\hat k) = -\hat i -\hat j +\hat k$

Since a is parallel to the line of intersection of the planes determined by$\hat i, \hat i +\hat j$ and $\hat i -\hat j,\hat i+\hat k$. Therefore,

$\vec a||\vec{n_1}×\vec{n_2}$

$⇒\vec a=λ(\vec{n_1}×\vec{n_2}=λ\begin{vmatrix}\hat i&\hat j&\hat k\\0&0&1\\-1&-1&1\end{vmatrix}=λ(\hat i-\hat j)$

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

$\cos θ=\frac{λ(\hat i-\hat j).(\hat i -2\hat j+2\hat k)}{\sqrt{λ^2+λ^2}\sqrt{1+4+4}}=\frac{λ(1+2)}{\sqrt{2}λ×3}=\frac{1}{\sqrt{2}}$

$⇒θ=\frac{π}{4}$