I'm stuck on this question:

Q: The vectors a, b, and c are of equal length l, and define the positions of the points A, B, and C relative to O. If OABC is a regular tetrahedron, find the distance of one vertex from the opposite face.

My attempt at the question was as follows:

Let the vector normal to the plane ABC be $\displaystyle

\hat n

$

then $\displaystyle

\hat n = \frac{{(a - c) \wedge (b - a)}}{{\left| {(a - c) \wedge (b - a)} \right|}}

$

distance d is the projection of c onto a, so $\displaystyle

d = (c.\hat n)\hat n

$

but if I substitute equation 1 into 2 it gives rise to a very complicated equation and I know I need to use the fact that a, b, and c are of length l.

Any help and pointers would be much appreciated.