Answer
$$3\sqrt 3$$
Work Step by Step
$\vec{r} (x,y) =x i+y j+(4-x-y) k$
$\implies \vec{r_x} \times \vec{r_y}=i+j+k$
$\implies |\vec{r_x} \times \vec{r_y}| =\sqrt {(1)^2+(1)^2+(1)^2}=\sqrt 3$
Now, $\iint_{S} F (x,y,z) \ d \theta=\int_{0}^{1} \int_{0}^{1} (4-x-y) \times \sqrt 3 \ dy \ dx \\=\sqrt 3\times \int_{0}^{1} [4y-xy-\dfrac{y^2}{2} ]_0^1 \ dx \\=\sqrt 3 \times \int_{0}^{1} 4(1-0)-x(1-0)-(\dfrac{1}{2}-0) \ dx \\=\sqrt 3 \times \int_{0}^{1} (\dfrac{7}{2} -x)] \ dx$
Now, we evaluate the integral
$I=\sqrt 3 \int_{0}^{1} (\dfrac{7}{2} -x)] dx= 3\sqrt 3$
