= Solution
Set $D=\mathbf a\cdot(\mathbf b\times\mathbf c)>0$. Since these vectors form a <basis>, write $\mathbf r=\alpha\mathbf a+\beta\mathbf b+\gamma\mathbf c$. The <reciprocal basis> gives
$$
\alpha=\frac{\mathbf r\cdot(\mathbf b\times\mathbf c)}D,\qquad
\beta=\frac{\mathbf r\cdot(\mathbf c\times\mathbf a)}D,\qquad
\gamma=\frac{\mathbf r\cdot(\mathbf a\times\mathbf b)}D.
$$
A point is in the closed <tetrahedron> precisely when it is a <convex combination> of the four vertices, with weights $1-\alpha-\beta-\gamma,\alpha,\beta,\gamma$. Thus the necessary and sufficient conditions are
$$
\boxed{\begin{gathered}
\mathbf r\cdot(\mathbf b\times\mathbf c)\geq0,\quad
\mathbf r\cdot(\mathbf c\times\mathbf a)\geq0,\quad
\mathbf r\cdot(\mathbf a\times\mathbf b)\geq0,\\
\mathbf r\cdot(\mathbf a\times\mathbf b+\mathbf b\times\mathbf c+\mathbf c\times\mathbf a)\leq D.
\end{gathered}}
$$
These four <half-spaces> are the four inward sides of its faces. This is <tetrahedron membership via scalar triple products>; making all inequalities strict describes the interior.
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