= Solution
Label $\mathbf a$ opposite $\mathbf x_0$, $\mathbf b$ opposite $\mathbf x_1$, $\mathbf c$ opposite $\mathbf x_2$ and $\mathbf d$ opposite $\mathbf x_3$. Since the <tetrahedron> is nondegenerate, its <scalar triple product> $T=\mathbf u_1\cdot(\mathbf u_2\times\mathbf u_3)$ is nonzero. Put $s=T/|T|$ to handle either <orientation of a vector space>. Then the required outward <vectors> are
$$
\boxed{\begin{aligned}
\mathbf a&=s(\mathbf u_2-\mathbf u_1)\times(\mathbf u_3-\mathbf u_1),\\
\mathbf b&=-s\,\mathbf u_2\times\mathbf u_3,\\
\mathbf c&=-s\,\mathbf u_3\times\mathbf u_1,\\
\mathbf d&=-s\,\mathbf u_1\times\mathbf u_2.
\end{aligned}}
$$
Each <cross product> has magnitude twice the area of its triangular face. The signs are fixed geometrically: $\mathbf b\cdot\mathbf u_1=-|T|<0$, so $\mathbf b$ points away from the opposite vertex; the same test with $\mathbf u_2$ and $\mathbf u_3$ handles $\mathbf c$ and $\mathbf d$. For the remaining face, the displacement toward its opposite vertex is $-\mathbf u_1$, and $\mathbf a\cdot(-\mathbf u_1)=-|T|<0$, again outward.
Expand the first <cross product>:
$$
\mathbf a=s(\mathbf u_2\times\mathbf u_3+\mathbf u_3\times\mathbf u_1+\mathbf u_1\times\mathbf u_2)
=-\mathbf b-\mathbf c-\mathbf d.
$$
Therefore
$$
\boxed{\mathbf a+\mathbf b+\mathbf c+\mathbf d=\mathbf0.}
$$
This is <closure of outward face area vectors>, with every face area multiplied by the same factor two. Omitting $s$ would orient all four faces inward when $T<0$.
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