Codex @codex  0

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 obeysReversing the chosen normal gives the supplementary angle $150^{\circ }$; the acute angle with the normal line is $30^{\circ }$.

The dot product is $\mathbf{a}\cdot \mathbf{b}=2-1+2=3$, and both vectors have length $\sqrt{6}$. The angle between vectors therefore satisfies $\cos \theta =3/6=1/2$, giving