Assume three points with three different coordinates in three-dimensional space. For example, let's assume points A, B and C have the following coordinates on the plane:
A = (1, 2, 3)
B = (-3, -5, 11)
C = (1, 3, -1)
Use this equation for a plane:
Ax + By + Cz = D
D is the distance from the origin (point 0,0,0). The x intercept is -D/A, the y intercept is -D/B, and the z intercept is -D/C. With these three intercepts, you can draw the plane in 3-dimensional space. Now, let's get specific about how to find these values from the coordinates from our example above.
Use a three-line matrix equal to zero to set-up your problem-- this looks like this:
[ x - x1, y - y1, z - z1],
[ x2 - x1, y2 - y1, z2 -z1],
[ x3 - x1, y3 - y1, z3 - z1]
= 0
Plug in the values you have from the original points (remember that x, y, and z are the intercepts):
[ x - 1, y - 2, z - 3],
[-3 - 1, -5 - 2, 1 - 3],
[ 1 - 1, 3 - 1, -1 - 3]
Simplifying this, we get:
[ x - 1, y - 2, z - 3],
[ -4, -7, -2],
[ 0, 2, -2]
Using these equations
A = (By - Ay)(Cz - Az) - (Cy - Ay)(Bz - Az)
B = (Bz - Az)(Cx - Ax) - (Cz - Az)(Bx - Ax)
C = (Bx - Ax)(Cy - Ay) - (Cx - Ax)(B7 - Ay)
D = a(Ax) + b(Ay) + c(Az)
to determine Ax + By + Cz + D = 0
yields a general equation of the plane of:
20x - 16y - 4z + 24 = 0