How to determine a point in $\mathbb{R}^3$ which is which is equidistant to three pairs of point?

Question

I have a three pairs of points namely : $p_{1,1}, p_{1,2}, p_{2,1}, p_{2,2}, p_{3,1}, p_{3,2}$, I want to find $p$ such that : $|p-p_{i,1}| = |p-p_{i,2}|$ for all $i$.

I know that these represent three planes, but I am not sure whether these planes do meet in a single point. My simple question is even if the three planes are not parallel do they always meet in a single point. Intuitively I can visualize three planes which are not parallel but don't meet at a single point but how do I know that mathematically ?

Answer 1

The planes described by the equations $z=0$, $z=y$, and $z=2y$ have the $x$-axis in common, so you may have infinitely many points. The planes described by the equations $z=0$, $z=y$, and $z=1-y$ have no points in common, but no two are parallel.

One way to determine whether the given planes will have a unique point of intersection is using linear algebra. Given the equations $a_1x+b_1y+c_1z=d_1$, $a_2x+b_2y+c_2z=d_2$, and $a_3x+b_3y+c_3z=d_3$ describing the planes, there is a unique solution to the system of equations, corresponding to the point where the planes meet, if and only if the coefficient matrix $$A= \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$$ has an inverse, in which case the solution is

$$\begin{bmatrix} x\\ y \\ z \end{bmatrix} =A^{-1} \begin{bmatrix} d_1\\ d_2 \\ d_3 \end{bmatrix}. $$