Convert a 2D point to 3D on a plane I have a 2D point and a 3D infinite plane(defined by a 3D point and its normal), I want to convert 2D point to 3D point by projected 2D point onto 3D plane surface.
I'm weak in math, I need a method to find the z-coordinate.
example:

 A: HINT:
If the plane equation is given  $ax+by+cz+d=0$
Normal vector of the plane  $$\overrightarrow{N}=(a,b,c)$$
Assume that the projection point on plane $A(x_1,y_1,z_1)$ that it is unknown and your aim is to find it. The point must satisfy the plane equation $ax+by+cz+d=0$
Given 2D point $B(x_2,y_2,0)$, I assumed that the point is on xy plane. Thus, I took $z=0$
$\overrightarrow{AB}$  must be parallel to $\overrightarrow{N}=(a,b,c)$ 
Thus
$$\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{0-z_1}{c}=k$$
 and also we know that 
$$ax_1+by_1+cz_1+d=0$$
You can solve the equations above and find the projection point $A(x_1,y_1,z_1)$  on the plane. 
A: Let the normal be
$N$
and the point on the plane
be $P$,
so the points in the plane are
$(A-P)\cdot N = 0$.
If your 2D point
is $T$
(with, perhaps, the third coordinate zero),
the line from $T$
in direction $N$ is
$T+rN$,
where $r$ is a real number.
If this is in the plane,
then
$(T+rN-P)\cdot N = 0$
or
$0
=(T-P)\cdot N + rN\cdot N
$
or
$rN \cdot N = (P-T)\cdot N$
or
$r
=\dfrac{(P-T)\cdot N}{N \cdot N}
$.
Your point in the plane,
as stated above,
is then
$T+rN$.
A: You can use the line-plane intersection algorithm given here (https://en.wikipedia.org/wiki/Line%E2%80%93plane_intersection).
Let $N$ be the normal of the plane and $p0$ be a point on the plane. 
Equation of a line: 
$$p = d*l + l0$$
Equation of a plane: 
$$(p-p0).N = 0$$
where $d$ is a scalar which defines the slope of the line
$l0$, $p0$ are lines and points on the plane 
$l$ is another point on the line which in your case will be the point projected on the plane. 
Solve for $d$,
$$d = (p0-l0).N/(l.N)$$
Your points on the plane will be 
$$d*l + l0$$
