Given three points P1, P2, P3, we can define two vectors in the plane determined by the points as V1 = P2 - P1 and V2 = P3 - P1. The vector n = V1 x V2 is then orthogonal (perpendicular) to both V1 and V2. If P is any point in the plane, then n is also orthogonal to P - P1 (and if P is not in the plane, then n and P - P1 are not orthogonal). So the equation of the plane can be written n · (P - P1) = 0. Let's apply this to an example.
Suppose P1 = (1,1,1), P2 = (2,3,1), P3 = (0,4,2). Then V1 = <1,2,0> and V2 = <-1,2,1>. We can find n as the determinant of
i j k
1 2 0
-1 2 1
which is <2,-1,4>. (Check that the dot products of this with V1 and V2 are both 0!) So our equation is:
<2,-1,4> · <x - 1, y - 1, z - 1> = 0
or 2x - y + 4z = 5
(It's a good thing to substitute each of the three points into this equation to confirm that you did everything right.)
Note that the coefficients of x, y and z are 2, -1 and 4, which is the normal vector we just found. A short cut for writing the equation of a plane is to write n1x + n2y + n3z (in this case, 2x-y+4z) on the left side, and substitute any of the three points to determine the value on the right side. So using P1 for example we have
2x - y + 4z = 2(1) -1 + 4(1) = 2-1+4 = 5.
How would you find the intersection of a line with a plane? Substitute the parametric representations for x, y and z from the line into the equation of the plane and solve for the value of the parameter. This is what we'll have to do when we start shooting projectiles at objects!