Find an equation of the plane through the three points given: p=(0,4,0),q=(2,1,2),r=(4,5,−2)

Pq × pr = [4, 12, 14]
so a suitable direction vector normal to the plane is [2, 6, 7]. The equation of the plane can be written as
.. 2x +6(y -4) +7z = 0


The cross product of two vectors in the plane will give a vector perpendicular to both, hence perpendicular to the plane. Given such a normal vector, the dot-product of that with any direction vector in the plane will be zero. (x, y, z) -p is such a direction vector in the plane.