Okay, well that's quite simple - you have the plane in 3D space, you
need to translate back into pixels.  You do that parametrically.  You
have the intersection (P) of the ray and the plane (VP) so convert the
offset of P relative to VP into a pair of x,y values in the VP frame of
reference (in the range 0-1) and multiply these values by the screen
dimensions to get to pixels.

To convert P,VP into an x,y cordinates:

Your plane is (A,B,C,D)

The point is P

x,y are the unknowns

   
    A    |            B
        | y
    --------P
       x

    D                C

The distance AP calculated using vector math (distance between A and P
is the magnitude of (A-P))

The angle between AP and AB can be calculated using the dot product.
The general rule is:

    r.q = |r| |q| cos(theta)

so we can get the angle from the dot product by rearanging:

    cos(theta) = r.q / (|r| |q|)

Once you have the angle of BAP and the length of AP you can find out the
 Y cordinate using trig:

    sin(theta) = y / |AP|
.:    y = sin(theta) * |AP|

You can calculate x in a similar way (angle PAD is 90 - BAP) so:

    sin(90-theta) = x / |AP|
.:    x = sin(90-theta) * |AP|

Thats how I figure it anyway, hopefully I haven't made any mistakes.
Don't forget to normalise the vectors before using the dot product to
get angles.

HTH

Peter.