Surface integral using Stokes' theorem $$
\int_\Gamma y\,dx+z\,dy+x\,dz
$$
when

$\Gamma$ $= \{ (x,y,z): x^2+y^2+z^2=9\}$ $\cap$ $x+y+z=0$

There's a theorem that states:
$\int_S(\nabla \times \vec F)\cdot d \vec S$= $\int_S(\nabla \times \vec F)\cdot \vec n\cdot d \vec S$

So, according to Stokes' theorem, I calculate $ \partial F $= $dx \wedge dy + dy\wedge dz + dz \wedge dx$.
I want the surface $S$ to be the plane $x+y+z=0$ inside the sphere, so I calculate its normal vector: $$\vec n = (\frac {1} {\sqrt3},\frac {1} {\sqrt3},\frac {1} {\sqrt3})$$
If I place $\partial F$ and $\vec n$ in the equation above, I will have $\frac {1} {\sqrt3}\int(x+y+z)\,dS$, and this is supposed to equal $0$ because $x+y+z=0$.

But the answer is not $0$, so  what am I doing wrong here?
Your mistake seems to have been in the computation of the curl of the vector field whose line integral you are evaluating.

Supposing $\vec{F}=\langle y,z,x\rangle$, then

$$\nabla\times\vec{F}=\langle -1,-1,-1\rangle.$$

Let $S= \{ (x,y,z): x^2+y^2+z^2=9\}$. This is the sphere of radius $3$ centered at the origin. The plane $x+y+z=0$ cuts this sphere into two hemispheres. Let $H$ be the hemisphere on the same side of the plane as the normal vector $\vec{n}=\langle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\rangle$.

$$\int_\Gamma ydx+zdy+xdz=\iint_{H}(\nabla\times\vec{F})\cdot\vec{n}\,\mathrm{d}S=-\sqrt{3}\iint_{H}\mathrm{d}S=-\sqrt{3}(2\pi9)\\=-18\sqrt{3}\pi,$$

where in the second to last step we've simply used the fact that the surface area of a hemisphere of radius $r$ is $2\pi r^2$, or half that of a sphere of radius $r$.
Your mistake seems to have been in the computation of the curl of the vector field whose line integral you are evaluating.

Supposing $\vec{F}=\langle y,z,x\rangle$, then

$$\nabla\times\vec{F}=\langle -1,-1,-1\rangle.$$

Let $S= \{ (x,y,z): x^2+y^2+z^2=9\}$. This is the sphere of radius $3$ centered at the origin. The plane $x+y+z=0$ cuts this sphere into two hemispheres. Let $H$ be the hemisphere on the same side of the plane as the normal vector $\vec{n}=\langle \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\rangle$.

$$\int_\Gamma ydx+zdy+xdz=\iint_{H}(\nabla\times\vec{F})\cdot\vec{n}\,\mathrm{d}S=-\sqrt{3}\iint_{H}\mathrm{d}S=-\sqrt{3}(2\pi9)\\=-18\sqrt{3}\pi,$$

where in the second to last step we've simply used the fact that the surface area of a hemisphere of radius $r$ is $2\pi r^2$, or half that of a sphere of radius $r$.