<s>[INST] How do I find the point equidistant from three points $(x, y, z)$ and belonging to the plane $x-y+3z=0?$ I struggle to find the point ${P}$ equidistant from the points ${A(1,1,1), B(2,0,1), C(0,0,2)}$ and belonging to the plane ${x-y+3z=0}$.
Any tips?
 [/INST] In $\Delta ABC$,
\begin{align*}
  a &= BC \\
  &= \sqrt{5} \\
  b &= CA \\
  &= \sqrt{3} \\
  c &= AB \\
  &= \sqrt{2} \\
  a^2 &= b^2+c^2 \\
  \angle A &= 90^{\circ} \\
  O &= \frac{B+C}{2} \tag{circumcentre of $\Delta ABC$} \\
  &= \left( 1,0,\frac{3}{2} \right) \\
  \vec{AB} \times \vec{AC} &=
  (1, -1, 0) \times (-1, -1, 1) \\
  &= (-1,-1,-2)
\end{align*}
Equation of axis of enveloping cone for circular section $ABC$ is 
$$\mathbf{r}=\left( 1,0,\frac{3}{2} \right)+t(-1,-1,-2)$$
Substitute into $x-y+3z=0$,
\begin{align*}
  (1-t)-(-t)+3\left( \frac{3}{2}-2t \right) &= 0 \\
  t &= \frac{11}{12} \\
  (x,y,z) &= \left( \frac{1}{12},-\frac{11}{12},-\frac{1}{3} \right)
\end{align*}
</s>