Find the shortest distance, d, from the point (1, 0, −6) to the plane x + y + z = 6

please help!

The distance is √3 / 3 (Answer). Here's one way to do it, using vectors ...

The points (6, 0, 0), (0, 6, 0), and (0, 0, 6) all lie in the plane, so the vectors <-6i + 6j> and <-6i + 6k> both lie in the plane, and their cross product <36i + 36j - 36k> = 36 is perpendicular to the plane. Converting that to a unit vector, we have (√3 / 3) as a unit vector perpendicular to the plane.

Using (1, 0, -6) and (6, 0, 0), the vector <5i + 6k> lies between the point and the plane, and the dot product (√3 / 3) <5i + 6k> • = (√3 / 3)(5 - 6) = -√3 / 3 is the projection of one vector on the other. The absolute value √3 / 3 is the distance from the point to the plane. (Answer)