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

Solution 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 $P$.

Solution 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.