= Solution
The face $PQR$ has edge vectors $\mathbf q-\mathbf p=(-1,1,1)$ and $\mathbf r-\mathbf p=(-2,-1,2)$. Their <cross product> is $(3,0,3)$, so choose the outward <normal vector> $\mathbf n=(1,0,1)$. The <angle between vectors> obeys
$$
\cos\theta=\frac{\mathbf p\cdot\mathbf n}{|\mathbf p||\mathbf n|}=\frac3{\sqrt6\sqrt2}=\frac{\sqrt3}2,\qquad\boxed{\theta=\frac\pi6=30^\circ.}
$$
Reversing the chosen normal gives the supplementary angle $150^\circ$; the acute angle with the normal line is $30^\circ$.
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