There are two methods: 1, the plane equation X-3Y+ 1 = 0, which is parallel to the plane through point (4,-1, 0), is (X-4)-3 (Y+ 1) = 0, that is, x-3y. The plane equation that the crossing point (4,-1, 0) is parallel to the plane 2y+z- 1 = 0 is 2 (y+ 1)+(z-0) = 0, that is, 2y+z+2 = 0. So the equation of the straight line l is X- 0) × (0,2, 1) = (-3,-1, 2), so the equation of the straight line l is: (x-4)/(-3) = (y+1) = z =-2y+/kloc. The parameter is y, so the direction vector of intersection line is (3, 1, -2), and the direction vector of straight line L is parallel to it, so the equation of straight line L is (x-4)/.