Find the third vertex of a triangle in 3D space

Question

I need to find the third vertex of an isosceles triangle in $3D$ space. What I know are coordinates of two other vertices and angles between sides of the triangle:

$|AB| = |AC|$. Coordinates of vertices $A$ and $B$ are known. What's more I know that this triangle is parallel to a plane given by its normal and a point that lies on it. I need to find coordinates of vertex $C$.

Answer 1

$$(2A-C-B)\cdot(B-C)=0$$

$${(A-B)\cdot(C-B)\over |A-B||C-B|} = \cos(a)$$

Let $C$ has coordinate $x_c,y_c,z_c$ and plug in to the above equation.