Find an equation of the plane that passes through the points p, q, and r. p(7, 2, 1), q(6, 3, 0), r(0, 0, 0)

x - 2y - 3z = 0

Step-by-step explanation:

The cross product of vectors rp and rq will give a vector that is normal to the plane:

... rp × rq = (-3, 6, 9)

Dividing this by -3 (to reduce it and make the x-coefficient positive) gives a normal vector to the plane of (1, -2, -3). Usint point r as a point on the plane, we find the constant in the formula to be zero. Hence, your equation can be written ...

... x -2y -3z = 0