Answer
$2$
Work Step by Step
The surface integral can be found as:
$dS= \sqrt {(f_x)^2 +(f_y)^2 +1} \ dy \ dx$
or, $dS=\sqrt {(-2)^2 +(-2)^2 +1} \ dy \ dx =3$
Now, $$\iint_{S} G (x,y,z) \ dS=\int_0^1 \int_0^{1-x} (2-x-y) (3) \ dy \ dx \\=3 \times \int_0^1 [2y-xy-\dfrac{y^2}{2}]_0^{1-x} \ dx \\=3 \times \int_0^1 [2-2x-x+x^2-\dfrac{(1-2x+x^2)}{2}] \ dx \\ =3 \times \int_0^1 (2-2x-x+x^2-\dfrac{1}{2}+x-\dfrac{x^2}{2}] \ dx \\= 3 \times \dfrac{2}{3} \\=2$$
