Prove that the area enclosed by a curve $C$ in a plane having normal $a\hat i+b\hat j+c\hat k$ is given by 
Stokes theorem:
$$\int\int (\nabla\times F)\cdot n\text{d}A=\oint_CF.\text{d}r$$
I took $$F=\frac{z}{b}\hat i+\frac{x}{c}\hat j+\frac{y}{a}\hat k$$ So that the LHS becomes $3A$. But the RHS does not match with the one given on the question.
Note that for $\vec F=\hat x(bz-cy)+\hat y (cx-az)+\hat z(ay-bx)$, we have
$$\oint_C \vec F(x,y,z)\cdot d\vec \ell=\int_C ((bz-cx)\,dx+(cx-az)\,dy+(at-bx)\,dz) \tag 1$$
Furthermore, note that $\nabla \times \vec F(x,y,z)=2\left(\hat x a+\hat yb+\hat zc\right)$ so that
$$\begin{align} \int_S \nabla \times \vec F(x,y,z) \cdot \hat n\,dS&=2\int_S \left(\hat x a+\hat yb+\hat zc\right)\cdot \frac{\left(\hat x a+\hat yb+\hat zc\right)}{\sqrt{a^2+b^2+c^2}}\,dS\\\\ &=2\int_S (1)\,dS\\\\ &=2\times\text{The area of}\,\,S \tag 2 \end{align}$$
Therefore, from Stokes's Theorem, $(1)$ and $(2)$ are equal and we can assert that
$$\frac12 \oint_C \vec F(x,y,z)\cdot d\vec \ell=\text{The area of}\,\,S$$