The acute angle that the vector $2\hat i-2\hat j +\hat k$ makes with the plane determined by the vectors $2\hat i+ 3\hat j-\hat k$ and $\hat i-\hat j+2\hat k$ is

$\sin^{-1}(\frac{1}{\sqrt{3}})$

Let $\vec a=2\hat i -2\hat j+\hat k, \vec b=2\hat i+3\hat j-\hat k$ and $\hat c=\hat i-\hat j+2\hat k$.

Let $\vec n$ be the normal to the plane. Then, $\vec n=\vec b×\vec c$.

Required angle θ is given by

$\sin θ=\frac{(\vec b×\vec c).\vec a}{|\vec b×\vec c||\vec a|}$

Now,

$\vec b×\vec c=\begin{vmatrix}\hat i&\hat j&\hat k\\2&3&-1\\1&-1&2\end{vmatrix}=5\hat i-5\hat j-5\hat k$

$∴(\vec b×\vec c)=\vec a=(5\hat i-5\hat j-5\hat k).(2\hat i -2\hat j+\hat k)=15$

$|\vec b×\vec c|=\sqrt{25+25+25}=5\sqrt{3}$ and, $|\vec a|=3$

Hence, $\sin θ=\frac{(\vec b×\vec c).\vec a}{|\vec b×\vec c||\vec a|}$

$⇒\sin θ=\frac{15}{15\sqrt{3}}=\frac{1}{\sqrt{3}}⇒θ=\sin^{-1}(\frac{1}{\sqrt{3}})$