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Let $\Omega$ be the interior of the tetrahedron, and take any constant vector field $\mathbf{q}$. Its divergence is zero, so the divergence theorem givesChoosing $\mathbf{q}$ to be each Cartesian basis vector proves ${\sum }_i{\mathbf{V}}_i=0$. This is the general fact that vector areas of a closed polyhedron sum to zero, specialized to four faces. The outward orientations are essential: the scalar face areas are all positive and do not sum to zero.

For the triangular face, take the cross product of two edges and divide by two: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, withThe vector areas of a closed polyhedron sum to zero, so the sum for the remaining three faces is

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.