I think you need to understand that, no, no one here is going to do your homework for you. You learn mathematics by doing mathematics, not by watching someone else do it for you.
Here's how I would approach this problem. The plane 2x+ 2y+ z= 6 can be written as z= 6- 2x- 2y or as the "position vector" .
The derivatives with respect to x and y are and . Those vectors are in the plane and their cross product, is perpendicular to the plane and its length is the "differential of area" for the plane. In general, the vector is the "vector differential of area" for the plane Ax+ By+ Cz= D.
Now, what about the orientation? You say "where is the outer normal to S". That would make sense for a closed surface but not for a plane! A plane has no "inside" or "outside". Which side of the plane is the normal to be on? Do you want this plane oriented by the "normal in the positive z direction" or in the "negative z direction"?
If positive z direction, take the dot product of with and integrate with respect to x and y. If the negative z direction, take the dot product of with and integrate (that will just change the sign on the answer).
The plane 2x+ 2y- z= 6, projected to the xy-plane, is 2x+ 2y= 6 or x+ y= 3. You can get the limits of integration from that.