That's either
vnormalize(B-A)*D+A
or
-vnormalize(B-A)*D+A
depending on which direction you're thinking of. The first
will either move towards or pass B when D is either smaller
or larger than the distance from A to B,
the second will move away from A and B on the line.

Now that's POV-Code, if you want the actual formula
to do it by hand, you can easily look up vnormalize in
the docs or search the web for normalizing vectors (which
is actually just dividing a vector by its length, which
can be calculated by taking the square-root of the addition
of each component (x, y, and z) squared...
That'd be:
square_root( x*x + y*y + z*z)

That isn't POV-Code, its Pseudo-Code to aid in understanding
the complicated sentence above. :-)

-- 
Tim Nikias v2.0
Homepage: http://www.digitaltwilight.de/no_lights


> If i have 2 points A <x1,y1,z1> and B <x2,y2,z2>, how do i find
coordinates
> of point C <x3,y3,z3> which lies on the line AB, distance D from point A?
>
> Thank you,
>
> Boris.
>


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