Normal vector of plane l? X+y-z = 3 is (1, 1,-1).

Normal vector of plane l? 4x+y-z = 3 equals (4, 1,-1).

Therefore, the direction vector n of the intersection of l? What about me?

= ( 1, 1, - 1)×(4, - 1, 5) = (4, -9, -5)

We can find a point on the intersection of l? What about me? ：

Let x = 0, we have y-z = 3, y-5z = -5, and the solution is z = 2 and y = 5.

Therefore, point P (0,5,2) is a point on the intersection line.

A is a given point (-2,3,3), and the vector AP = (2 2,2,-1).

Vector in the normal direction of the required plane.

= N crossing time AP

= (4, -9, -5)×(2, 2, - 1) = ( 19, -6, 26)

Sub (19,-6,26) and given point (-2,3,3) are transformed into ax+by+cz = d, and we get

- 38 - 18 + 78 = d，d = 22

∴The equation of the required plane is 19x-6y+26z = 22 (Ans).