Find the third vertex of a triangle in 3D space

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 $$C$$.

• It seems to me, there is a whole circle of points that satisfy the condition in 3D. Even if you define a plane the triangle lies on, there still will be two possible solutions. Aug 20, 2020 at 11:55
• There is not a unique solution but a an infinite set of solutions : a circle (rotate the triangle around line AB). Aug 20, 2020 at 11:56
• I know that this triangle has to be prallel to a plane given by its normal and a point that lies on it [added also to the question] Aug 20, 2020 at 12:04

$$(2A-C-B)\cdot(B-C)=0$$
$${(A-B)\cdot(C-B)\over |A-B||C-B|} = \cos(a)$$
Let $$C$$ has coordinate $$x_c,y_c,z_c$$ and plug in to the above equation.