= Solution
Noncoplanarity makes the <scalar triple product> $\Delta$ nonzero, so the three displayed vectors form a <reciprocal basis>. Their scalar products obey
$$
\mathbf A\cdot\mathbf a=\mathbf B\cdot\mathbf b=\mathbf C\cdot\mathbf c=1,
$$
with all six cross-pairings zero. Therefore, assuming the nonzero $\alpha,\beta,\gamma$ needed to define the three given points,
$$
\mathbf X\cdot\frac{\mathbf a}{\alpha}=\mathbf X\cdot\frac{\mathbf b}{\beta}=\mathbf X\cdot\frac{\mathbf c}{\gamma}=1.
$$
Every difference of two of these points is <orthogonal> to $\mathbf X$. The two independent differences span their <plane>, so \b[$\mathbf X$ is a normal to the plane], whose equation is $\mathbf X\cdot\mathbf r=1$. Also $\mathbf X\ne0$, because its scalar product with each specified point is $1$.
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