= Solution
For the triangular face, take the <cross product> of two edges and divide by two:
$$
\frac12\overrightarrow{PQ}\times\overrightarrow{PR}
=\frac12(-1,1,0)\times(-2,0,1)=\frac12(1,1,2).
$$
The origin lies on the opposite side from this normal, since $(1,1,2)\cdot\overrightarrow{PO}=-3$. Thus this is already the outward <vector area>, with
$$
\boxed{A=\frac{\sqrt6}{2},\qquad\hat{\mathbf n}=\frac{(1,1,2)}{\sqrt6},\qquad
A\hat{\mathbf n}=\left(\frac12,\frac12,1\right)}.
$$
The <vector areas of a closed polyhedron sum to zero>, so the sum for the remaining three faces is
$$
\boxed{\left(-\frac12,-\frac12,-1\right)}.
$$
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-2-tetrahedron.png]
{title=Tetrahedron with its highlighted face PQR and outward unit normal}
Their individual outward <vector areas> are $(0,0,-3/2)$, $(1/2,-1,1)$, and $(-1,1/2,-1/2)$, which directly check that sum.
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