How do you find an equation of the plane through the points (2, 1, 2), (3, 8, 6), and (2, 3, 1)?

To find the equation of a plane given three points, we can use the formula derived from the determinant of three vectors formed by these points.

1. **Identify the points**: Let the points be A(2, 1, 2), B(3, 8, 6), C(2, 3, 1).

2. **Form vectors**: Create two vectors from these points:

• AB = B – A = (3 – 2, 8 – 1, 6 – 2) = (1, 7, 4)
• AC = C – A = (2 – 2, 3 – 1, 1 – 2) = (0, 2, -1)

3. **Cross product**: The normal vector of the plane can be found by taking the cross product of vectors AB and AC:

n = AB × AC

Using the determinant format, we have:

| i    j    k |
| 1    7    4 |
| 0    2   -1 |

Calculating this gives us:

n = i(7 * -1 – 4 * 2) – j(1 * -1 – 4 * 0) + k(1 * 2 – 7 * 0) = i(-15) + j(1) + k(2) = (-15, 1, 2)

4. **Equation of the plane**: The general equation of a plane is given by:

Ax + By + Cz = D

Substituting the normal vector (-15, 1, 2) and point A(2, 1, 2) into the equation:

-15(x – 2) + 1(y – 1) + 2(z – 2) = 0

This simplifies to:

-15x + 1y + 2z = -15*2 + 1*1 + 2*2 = -30 + 1 + 4 = -25

So the equation of the plane is:

-15x + y + 2z = -25