proof that a line XO, where X is equidistant from A,B,C is perpendicular to the plane P

Question

Let $X$ be equidistant from points $A,B,C$, where $A,B,C$ lie in the same plane $P$ and form a (non-degenerate) triangle and $X$ is not in the plane $P$. Does it follow that $XO$ is perpendicular to $P$, where $O$ is the circumcenter of triangle ABC? If not, does it follow if $ABC$ is isoceles?

I tried proving the claim when ABC is isoceles using vectors, but I got stuck at some point. Take O to be the origin. Then letting $\vec{X} = \vec{OX}$, etc. we have $(\vec{X} - \vec{B})^2 = (\vec{X}-\vec{A})^2 = (\vec{X}-\vec{C})^2$ and using the fact that the magnitudes of the vectors $\vec{A},\vec{B}, \vec{C}$ are all equal we get that $\vec{X}\cdot \vec{B} = \vec{X}\cdot \vec{A} = \vec{X}\cdot \vec{C}$. In the isoceles case, suppose WLOG that $AC = BC$ so that $(\vec{C}-\vec{A})^2 = (\vec{C}-\vec{B})^2$, from which we obtain $\vec{A}\cdot \vec{C} = \vec{B}\cdot \vec{C}$. However, I'm not sure how to prove that $\vec{X}\cdot \vec{B} = \vec{0}$ in this case.

Answer 1

Your proof is almost good. In fact, from $\vec{X}\cdot\vec{A}=\vec{X}\cdot\vec{B}=\vec{X}\cdot\vec{C}$ you can conclude that $\vec{X}\cdot(\vec{B}-\vec{A})=\vec{X}\cdot\vec{AB}=0$ and $\vec{X}\cdot(\vec{C}-\vec{A})=\vec{X}\cdot\vec{AC}=0$, thus $\vec{X}=\vec{OX}$ is perpendicular to both $\vec{AB}$ and $\vec{AC}$, i.e. it is perpendicular to the whole plane $P$.

Answer 2

Yes.

$X$ is equidistant from $A,B$, so $X$ is in the plane $\pi_1$ containing the midpoint of $AB$ and perpendicular to $AB$. Similarly it is in the plane $\pi_2$ containing the midpoint of $BC$ and perpendicular to $BC$. So $X$ is on the line $\ell=\pi_1\cap \pi_2$ which is perpendicular to $AB, BC$, hence to the plane $P$. Both $\pi_1,\pi_2$ contain $O$ since $O$ is equidistant from $A,B, C$, hence $\ell=OX$. "Isosceles" is not used in this proof.