Find the Cartesian equation of a plane containing the points A(1, 7, -2) B(4, -3, 2) and C(7, 8, 9).

It can be shown that the equation of any plane can be given by r.n=a.n, where r = xi + yj + zk, n is a direction vector normal to the plane, and a is any point vector on the plane. a can be found easily (three points are already given in the question for us to choose from – in this case we’ll choose point A for simplicity). As such, the bulk of the work comes in finding n.
As n is a direction vector normal the plane, this can be found by evaluating the cross product of any two direction vectors lying on the plane. We can use the three given points to find both vectors: AB and AC, where AB = B – A and AC = C – A. In our case, AB = 3i – 10j + 4k and AC = 6i + j + 11k. Carrying out the cross product operation on these gives n = -114i – 9j + 63k.
Finally, we substitute these values into r.n=a.n giving (xi + yj + zk).(-114i – 9j + 63k) = (i + 7j – 2k).(-114i – 9j + 63k), which simplifies to -114x – 9y + 63z = -303.