Answer
$-\dfrac{1}{2}$
Work Step by Step
Stoke's Theorem states that $\iint_S curl F\cdot dS=\int_C F \cdot dr$
The equation of the plane with the given intercepts is given as:
$\dfrac{x}{1}+\dfrac{y}{1}+\dfrac{z}{1}=1$
This gives: $z=1-x-y$
Use definition of curl $F=\nabla \times F$
Consider $F=ai+bj+ck$
That is, curl $F=(\dfrac{\partial c}{\partial x}-\dfrac{\partial b}{\partial z})\hat{i}+(\dfrac{\partial a}{\partial z}-\dfrac{\partial c}{\partial x})\hat{j}+(\dfrac{\partial b}{\partial x}-\dfrac{\partial a}{\partial y})\hat{k}$
So, we have : curl $F=-yi-zj-xk$
$\iint_S curl F\cdot dS=\iint_D-y-z-x dA=-\iint_D dA$
Here, $\iint_D dA$ represents the area of the triangle equal to $\dfrac{1}{2}(Height)(Base)=\dfrac{1}{2}(1)(1)=\dfrac{1}{2}$
Hence, $\iint_S curl F\cdot dS=\int_C F \cdot dr=-\iint_D dA=-\dfrac{1}{2}$
