project a point in 3D on a given plane A point in a 3D space is given as $ P(x,y,z) $. I want to find the position of this point projected parallel to the normal on a plane Q defined by $3$ non-collinear points $ Q1(x1,y1,z1), Q2(x2,y2,z2)\; and \;Q3(x3,y3,z3) $.
How to derive this point of projection?
I saw this entry on Wikipedia but I cannot understand it. 
I'd prefer an answer that gives the derivation of a $4$-ary transformation function $f$ such that 
$ f \; (P, Q1, Q2, Q3) = P' $
where P' is a point in 2D with co-ordinates as $ P'(x',y') $ such that P' is the required projection of P on the plane Q.

Alternatively ...
Define $U$, $V$ as in my first answer. Let $N$ be a unit vector normal to the plane, which you can obtain by unitizing the cross product $U \times V$.
Then the projected point $R$ is given by
$$
R = P - \left[(P-Q_1) \cdot N\right]*N
$$
This is essentially the same answer provided by zuggg. I wrote it before seeing his, and decided to keep it since I think it's easier to read. You can decide which one you like better, and how to give credit (if any).
