8 Marks 5. Find a scalar equation of the plane containing (1, 2, 1 ) and containing the line of intersection between the planes & + 2y + 4z = 4 and 3x + 2y + 2z = 12. Solution: First we find the line of intersection: 2 4 4 2 2 12 row reduces to 0 2 0 1 O so the line of intersection is y = +t -5 = 0Q + Ed . Z Now, PQ = [3 -2 -1] and d = [2 -5 2]" are in the plane, so the normal to the plane is'n = d x PQ = [9 8 11]1 So a scalar equation of the plane is 9(x - 4) + 8(y - 0) + 11(z - 0) = 0, i.e. 9r + 8y + 11z = 36