Applying Green's Theorem I'm studying for a grad-school preliminary exam, and came across this problem which I am unable to solve. 

Let $C$ be a closed curve in the plane $ax + by + cz = 0$ (where $a,b,c \in \mathbb{R}$ are not all zero $0$), enclosing a region with area $S$.  Evaluate 
$$ I:= \oint\limits_C \left| \begin{array}{ccc} 
dx & dy & dz \\
a  & b  & c  \\
x  & y  & z \end{array} \right|
$$
where the integral along $C$ is counterclockwise relative to the normal direction $(a,b,c)$ to the plane.

Expanding the determinant yields $$
I = \oint\limits_C (bz - cy)dx + (cx - az)dy + (ay - bx)dz. $$
If we call $\Omega$ the region bounded by $C$, an application of Green's Theorem (or Stokes Theorem, if you prefer) yields $$
I = 2\iint\limits_\Omega a dydz + bdzdx + c dxdy.
$$
From here, I'm not sure what to do; I suspect that it's a simple solution and that I've just forgotten my vector calculus.  Anyway, any help is appreciated.

EDIT: Substitute in $z = \frac{1}{c}\left(-ax - by \right)$ and then applying Green's Theorem doesn't bring me any closer.  Making such a substitution yields $$ I = \frac{a^2 + b^2 + c^2}{c} \iint_\Omega dxdy $$
If $c \neq 0$ as required to make such a substitution, it is not the case that $\iint dxdy = S$.

SECOND EDIT:  I've solved the problem, making use of the following version of Stokes' theorem: $$ \iint_\Omega (\nabla \times \mathbf{F} ) \cdot \mathbf{n} dS = \oint_{\partial \Omega} \mathbf{F} \cdot d\mathbf{r}. $$ 
The above can be written as \begin{align*} I &= \oint\limits_C (bz - cy,cx - az,ay - bx)\cdot d\mathbf{r} \\
&= \iint_\Omega (\nabla \times (bz - cy,cx - az,ay - bx)) \cdot \left(\frac{1}{\sqrt{a^2 + b^2 + c^2}}(a,b,c)\right) dS  \\
&=\iint_\Omega (2a,2b,2c) \cdot \left(\frac{1}{\sqrt{a^2 + b^2 + c^2}}(a,b,c)\right) dS  \\
&=  2 \sqrt{a^2 + b^2 + c^2} \iint_\Omega dS  \\
&= 2\sqrt{a^2 + b^2 + c^2} (\text{Area of } \Omega) \end{align*}

Use the plane equation as 
$$ ax + by + cz = 0  \implies z = -\frac{a}{c}x -\frac{b}{c}y  \implies dz= -\frac{a}{c}dx - \frac{b}{c}dy $$
and substitute for $z$ and $dz$ in the integral
$$ \oint\limits_C (bz - cy)dx + (cx - az)dy + (ay - bx)dz. $$
Then Green's theorem is straightforward.
