Answer
$$ \dfrac{ -\sqrt 2}{2}$$
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 {(-1)^2+0^2+1} \ dz \ dy =\sqrt 2 $
Now, $$I=\iint_{S} G (x,y,z) \ dS=\int_0^1 \int_{0}^{1} (1-2y-z) \sqrt {2} \ dz \ dy \\=\sqrt 2 \times \int_0^1 [z-2yz -\dfrac{z^2}{2}]_0^1 \ dy \\=\sqrt 2 \times \int_0^1 [1-2y(1) -\dfrac{1}{2}(1^2-0)] \ dy \\=(\sqrt 2) \int_0^1 [\dfrac{1}{2}-2y] \ dy \\= \sqrt 2 \times \dfrac{1}{2} (y)]_0^1 - \dfrac{1}{2} \times [y^2]_0^1 \\= \dfrac{ -\sqrt 2}{2}$$
