Given three coordinates, which could be $A=(7,3)$, $B=(2,4)$, $C=(k,-2)$ I want to find the values of $k$ that make a right angle diagram out of the three points.
So I initially was thinking to find $M_{AC}=-1/M_{BC}$. In this way however I will find only the right angle that happen to be at $BAC$ and $ABC$ but not $ACB$. I can not think of a better method and more concise (probably pythagoras?) that allow me to find all three? Or even, is my method correct?
Your method is fine. You have to do it three times, once for each angle being the right angle. I would use Pythagoras, so to have $A$ the right angle you need $26+(k-7)^2+25=(k-2)^2+36$. It seems simpler to me, but again takes three tries.