Find the equation of the plane determined by the points A (3, - 1, 2), B (5, 2, 4) and C (–1, –1, 6). Also find the distance of the point P (6, 5, 9) from the plane?

Answer 1

The sophistication of this question indicates that you have some knowledge of determinants:

The equation of the plane is

| x-3 y+1 z-2|

| 2 3 3 | = 0

| 4 0 4 |

Now the distance from a point (x, y, z) to a plane Ax+ By + Cz = D is

| (Ax + By + Cz - D) / √(A² + B² + C ²) |