 Solution

Noncoplanarity makes the scalar triple product $\Delta$ nonzero, so the three displayed vectors form a reciprocal basis. Their scalar products obeywith all six cross-pairings zero. Therefore, assuming the nonzero $\alpha ,\beta ,\gamma$ needed to define the three given points,Every difference of two of these points is orthogonal to $\mathbf{X}$. The two independent differences span their plane, so $\mathbf{X}$ is a normal to the plane, whose equation is $\mathbf{X}\cdot \mathbf{r}=1$. Also $\mathbf{X}\ne 0$, because its scalar product with each specified point is $1$.