Thanks for the replies. You'll have to bear with me because for all the complex procedural programming concepts I understand, I struggle with mathematics.

If we take a step back; what I am trying to do is parallel a straight line segment. Therefore the point C will be one side of the paralleled line and I will create another point adjacent to B as the other point of that same paralleled line. Can you provide some advice on how I can parallel that straight line segment?

The problem is unsolvable with only two initial points, as this will produce a wheel of possible positions of C around A, remember this is 3D, so there exists infinite directions orthogonal to the vector (AB).

If I remove the 3d aspect, can you provide advice on how I can find point C's location and/or parallel a straight line segment in 2d? Is it simply a case of adding the offset(y value) to vector A? For eg:

Vector3 c = a + new Vector(y, 0, 0);  // y is the triangle length value from the OP picture
vector3 d = b + new Vector(y, 0, 0);

For the following line; if I use the above code wont that produce this result though?

So I have to consider the 2d rotation aswell right?