How do I find the point equidistant from three points $(x, y, z)$ and belonging to the plane $x-y+3z=0?$

Question

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?

Answer 1

Points equidistant from $A$, $B$ and $C$ lie along a line. This line is the intersection of various planes that bisect the line segments joining pairs of these three points. From there, you just need to find the point of intersection of this line with the given plane, which comes down to solving a system of linear equations.

Answer 2

Hint:

1)find the bisector plane of the segment $AB$, i.e the plane that passes thorough the point $M=((x_A+x_B)/2,(y_A+y_B)/2,(z_A+z_B)/2)=(3/2,1/2,1)$ and orthogonal the the vector $\vec{AB}=(1,-1,0)$. It should be :

$$\alpha) \qquad x-y-1=0 $$

2)find the bisector plane of the segment $CB$ (same way). It should be :

$$\beta) \qquad 4x-2z-1=0 $$

3) the point that you want is the intersection of the planes $\alpha,\beta$ and the given plane $\gamma $ of equation $x-y+3z=0$

$$ \begin{cases} x-y-1=0\\ 4x-2z-1=0\\ x-y+3z=0 \end{cases} $$

Answer 3

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*}