Find the distance from the point Q(8, - 2,4) to the plane 3x -y + 4z = 12. (Note that if P is a point on the plane ax + by + cz = d, then the distance from any point Q to the plane equals the length of the orthogonal projection of PQ onto a vector n = (a,b,c) normal to the plane. |PQ . n This can be expressed as d = The distance is. (Type an exact answer, using radicals as needed.)