Answer
$-\dfrac{1}{2}$
Work Step by Step
Consider the Stoke's Theorem states: $\iint_S curl F\cdot dS=\int_C F \cdot dr$
Here, we have
$\dfrac{x}{1}+\dfrac{y}{1}+\dfrac{z}{1}=1$
This is the equation of the plane with the intercepts.
Then, $z=1-x-y$
Let us consider $F=ai+bj+ck$
and $curl(F)=(\dfrac{\partial c}{\partial x}-\dfrac{\partial b}{\partial z})\hat{i}+(\dfrac{\partial a}{\partial z}-\dfrac{\partial c}{\partial x})\hat{j}+(\dfrac{\partial b}{\partial x}-\dfrac{\partial a}{\partial y})\hat{k}$
or, $ curl(F)=-yi-zj-xk$
and $\iint_S curl F\cdot dS=\iint_D-y-z-x dA=-\iint_D dA$
Here, $\iint_D dA$: shows the area of a triangle which is equal to $=\dfrac{1}{2}(1)(1)=\dfrac{1}{2}$
Thus, $\iint_S curl F\cdot dS=\int_C F \cdot dr=-\iint_D dA$
or, $=-\dfrac{1}{2}$
