= Solution
Compute twice the four opposite face areas and their weighted vertex average:
$$
\begin{aligned}
w_A&\leftarrow\|(C-B)\times(D-B)\|,\\
w_B&\leftarrow\|(C-A)\times(D-A)\|,\\
w_C&\leftarrow\|(B-A)\times(D-A)\|,\\
w_D&\leftarrow\|(B-A)\times(C-A)\|,\\
w&\leftarrow w_A+w_B+w_C+w_D,\\
I&\leftarrow(w_AA+w_BB+w_CC+w_DD)/w.
\end{aligned}
$$
These positive weights make $I$ an interior <convex combination>. To verify the <incenter of a tetrahedron>, put $\Delta=|(B-A)\cdot((C-A)\times(D-A))|$. The altitude from $A$ to its opposite face is $\Delta/w_A$. Signed distance to that face is affine and vanishes at the other three vertices, so $I$ has distance $(w_A/w)(\Delta/w_A)=\Delta/w$. The same argument works for every face. Hence
$$
\boxed{I\text{ is the incentre},\qquad r_{\rm in}=\Delta/w.}
$$
The sphere of radius $r_{\rm in}$ centered at $I$ is tangent to all four faces. Equality of the four distances conversely forces these face-area weights, proving uniqueness.
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