Take the equation of the plane and the equation of the line
(parametrically), and solve them as a system of equations.

For instance...

plane: 2x +4y +5z = 0
line:
x=t
y=2t
z=3t+5

then you plug in x y and z into the plane's equation, and solve for t...

2*t + 4*2*t + 5*3*t + 5*5 = 0
25*t = -25
t = -1

then, once you know t, just plug it in to the line's equation to find the
point of intersection:

x = -1
y = 2*-1 = -2
z = 3*-1 + 5 = 5 - 3 = 2

so the intersection is at <-1,-2,2>. Unless I messed something up, which is
entirely possible...

 - Slime
[ http://www.slimeland.com/ ]

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