Here's how to do it directly.
Flux = $\displaystyle \int \int_{S} F \cdot N \ \mathrm{d} S = \int \int_{D} F \cdot (T_{x} \times T_{y}) \ \mathrm{d}x \ \mathrm{d}y$
D is the region in the xy-plane bounded by the positive x and y axes and the line y = 1-x
Let $\displaystyle g(x,y) = z = 1-x-y $
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The surface can be parametrized by $\displaystyle \Phi(x,y) = \Big (x,y,g(x,y)\Big) = (x,y,1-x-y) $
$\displaystyle T_{x} = \frac {\partial \Phi}{\partial x} = (1,0,-1) $
$\displaystyle T_{y} = \frac {\partial \Phi}{\partial y} = (0,1,-1) $
$\displaystyle T_{x} \times T_{y} = (1,1,0) $
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The problem wants the normal vector pointing in the other direction, or $\displaystyle (-1,-1,0) $
but then $\displaystyle F \cdot (T_{x} \times T{y}) = 0 $? Is that right? Anybody?
D is the region in the xy-plane bounded by the positive x and y axes and the line y = 1-x
The final final integral is then $\displaystyle \int_{0}^{1} \int_{0}^{1-x} 0 \ \mathrm{d}y \ \mathrm{d}x $ which of course is just zero.