Plane through 3 given Points:

We need to determine the normal vector, n, for the desired plane. Since the vectors AB and AC lie in the plane, their cross product, which is perpendicular to the plane of AB and AC, can be used as the desired normal, n. Then we can use the scalar equation of the plane:

a*[x - x_A] + b*[y - y_A] + c*[z - z_A] = 0, where A(x_A, y_A, z_A) is a reference point in the plane and a, b, c are the components of the normal vector.

First we determine the vectors

AB = [x_B - x_A]i + [y_B - y_A]j + [z_B - z_A]k

= [(-4) - 1]i + [2 - (-6)]j + [(-5) - 0]k = -5i + 8j - 5k

AC = [x_C - x_A]i + [y_C - y_A]j + [z_C - z_A]k

= [(-2) - 1]i + [4 - (-6)]j + [1 - 0]k = -3i + 10j + k

Now form the cross product

, or

Using the components of this normal in the equation for the plane, we get (using point A's coordinates):

58*[x - 1] + 20*[y - (-6)] - 26*[z - 0] = 0

Simplifying, we get

58x + 20y - 26z = -62.

As a check on our work, we can substitute each of the three points into this equation to make sure that they satisfy it:

A(1, -6, 0): 58*1 + 20*(-6) - 26*0 = -62

B(-4, 2, 5): 58*(-4) + 20*2 - 26*(-5) = -62

C(-2, 4, 1): 58*(-2) + 20*4 - 26*1 = -62