Use Stokes' theorem to show that the line integrals have the given value Problem

Prove using Stokes' theorem that 
$$\int_C y dx +z dy + x dz = \pi a^2 3^.5,$$ where $C$ is the curve of intersection of the sphere $x^2+y^2+z^2=a^2$ and the plane $x+y+z=0$.

Attempt

$$\int_C y dx +z dy + x dz= \iint_S \nabla × F.n dS,$$ where $n$ is the normal to the surface $S$. $S$ is the ellipse formed by the intersection.
$F=(y,z,x)$
Now $z=-x-y$. 

Therefore $$n = \frac{\frac{\partial z}{\partial x} i+ \frac{\partial z}{\partial y} j-k}{|\frac{\partial z}{\partial x} i+ \frac{\partial z}{\partial y} j-k|}
=\frac{-i-j-k}{3^.5}$$

$$dS =\left(1+\frac{\partial z}{\partial x}^2 + \frac{\partial z}{\partial y}^2\right)^{\frac{1}{2}}$$
Therefore 
$$\iint_S (y,z,x).n dS=3 \iint dx dy$$

How to proceed after this ?

NB. This question is from Apostol's Calculus Volume 2,c Chapter 12, Section 12, Question 5, page 442 (2nd Edition).
You don't need to work out what $dS$ is.  The curl of $F$ is $\langle -1,-1,-1\rangle$, and the unit normal vector is the same thing, but divided by $\sqrt{3}$, so your surface integral is

$$\iint_S \frac{3}{\sqrt{3}} \; dS =\sqrt{3}\times \mbox{area of }S. $$

Since $S$ is a disc of radius $a$, you get 

$$\sqrt{3}\pi a^2.$$
S is the unit disk in the xy-plane and $\int_S\int dxdy$ is simply its area.