Answer
$-1$
Work Step by Step
Stokes' Theorem states that $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr$
where, $C$ corresponds to the boundary of the surface oriented counter-clockwise.
Now, $\iint_{S} curl \space F \cdot dS=\iint_{D} -(-2z)(-1) -(-2x) (-1) -2y dA$
Consider $z=1-x-y$
$\iint_{S} curl \space F \cdot dS=\iint_{D} -2+2x+2y-2x-2y dA$
Since, $D$ is the triangle formed by the vertices $(0,0); (1,0)$ and $0,1)$
Therefore, $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr =-2 \times \iint_{D} dA=-(2)(\dfrac{1}{2})=-1$
