Find the equation of the plane that passes through the points (5,1,3) and (2,-2,1) and is perpendicular to the plane 2x+y-z=4. Solution The equation of any plane passing through the point ( 5,1,3) is a ( x -5) + b ( y - 1 ) + c ( z - 3 ) = 0 .....(i) Since the point ( 2, -2, 1 ) lies on the plane (i) 3a + 3b + 2c = 0 ....(ii) Again the plane (i) is perpendicular to the plane 2x + y - z + 4 =0 2a + b -c =0 ....(iii) From (ii) and (iii), by cross multiplication method, we get, a/5 = -b/7 = c/3 Hence the required equation of the plane is 5(x - 5) -7(y -1) + 3(z -3) =0 ....[ From (i)] or 5x - 7y + 3z = 27.