If the triangle A,B,C has a right angle at the corner A, what is x?

Question

You are given the points $A(−2,−2,3)$, $B(4,9,−10)$ and $C(−2,x,1)$ in $ℝ^3$. If $\Delta ABC$ has a right angle at the corner $A$, what is $x$?

I tried finding the vectors $\vec{AB}$, $\vec{AC}$ and $\vec{CB}$. I did pythagorean theorem with the vectors which resulted in me finding $x=6.81$ however, I am not sure if I did it right.

Answer 1

$$\vec{AB}(4-(-2),9-(-2),-10-3)$$ or $$\vec{AB}(6,11,-13).$$ $$\vec{AC}(0,x+2,-2).$$ By the given $$\vec{AB}\cdot\vec{AC}=0.$$ Thus, $$6\cdot0+11(x+2)+(-13)(-2)=0,$$ which gives $$x=-\frac{48}{11}.$$

Answer 2

If $\Delta ABC$ has right angle at $A$, then $\vec{AB} \bot \vec{AC}$, which also means that scalar product of two vectors has to be $0$, that is $$\vec{AB} \cdot \vec{AC} = 0$$ $$\implies (6,11,-13) \cdot (0,x+2,-2) = 0$$ $$\implies 6\cdot0+11(x+2)+(-13)\cdot(-2) = 0$$ $$\implies 0+11x+48 = 0$$ $$\implies x = -\frac{48}{11}$$