Proving $AB$, $AC$ perpendicular to each other when vectors are median

Question

I want to prove this claim:

Triangle $ABC$ with $A(2,4,6),B(6,2,2),C(0,0,0)$ median, $AC$ and $BC$ perpendicular to each other.

What I did is to the $AB$, $BC$ make a dot product and thought it will be zero, but no result.
there is something I miss here.
Any suggestions?
Thanks!

Answer 1

The median to $AC$ is $$\frac{A+C}{2}-B=(-5,0,-1), $$ and the median to $BC$ is $$\frac{B+C}{2}-A=(1,-3,-5) .$$ Their dot product is indeed zero.