The integral $\int_C F\,dr$ is a line integral and does not require the normal to a surface when evaluated directly. Use an arbitrary parametrization $[a,b]\ni t\mapsto r(t)$ of the curve $C$ and trust your calculus:
$$
\int_C F\,dr=\int_a^b F(r(t))\cdot\frac{dr}{dt}\,dt\,.
$$
The integrand is a dot product of two vectors. Can you proceed? Once you calculated that you can match it with the surface integral of curl $F$. This will bring you in to a position to self-answer the original question. One more thing: how many surfaces can you choose to perform that exercise? Obviously you choose one that is as simple as possible.
To approach the surface integral directly you can parametrize the surface $S$ by
$$
[c,d]\times[e,f]\ni (u,v)\mapsto s(u,v)\,.
$$
The normal vector to $S$ is $s_u\times s_v$ (cross product of partial derivatives). The surface element is
$$
dS=\|s_u\times s_u\|\,du\,dv
$$
and we use it to calculate the total area of the surface. See a concrete example here.
The unit normal is of course
$$
\mathbf{n}=\frac{s_u\times s_u}{\|s_u\times s_u\|}\,.
$$
Stokes says
$$
\int_C F\,dr=\int_S {\rm curl\,}F\cdot\mathbf{n}\,dS\,.
$$
(Take an $F$ that satisfies ${\rm curl\, F}=\mathbf{n}\,.$ Its flux is constant equal to one on $S$ and the RHS must therefore be the total area of $S\,.$) Put back the definitions to get
$$\tag{1}
\int_a^b F(r(t))\cdot\frac{dr}{dt}\,dt=\int_c^d\int_e^f{\rm curl}F(s(u,v))\cdot (s_u\times s_v)\,du\,dv\,.
$$
The rules to remember are the following:
In formula (1) you dont't normalize the normal vector $(s_u\times s_v)$ because $\|s_u\times s_v\|$ has cancelled out and $s_u\times s_v$ needs to scale with the parametrization to ensure independence of that integral under a re-parametrization (change of variables).
When your problem has enough symmetry to see that curl$\,F\cdot\mathbf{n}$ is constant on $S$ then you just have to figure out the surface of $S$ and multiply it with that constant. In that case you normalize $\mathbf{n}$ of course.
In your example
$$
F=\begin{pmatrix} 2y\\-x\\y-x\end{pmatrix}\,,\quad S=\{x+y+z=6\}
$$
the second method is smarter:
curl$\,F\cdot\mathbf{n}=\color{red}{\frac{1}{\sqrt{3}}}\left(\begin{smallmatrix}1\\1\\-3\end{smallmatrix}\right)\cdot\left(\begin{smallmatrix}1\\1\\1\end{smallmatrix}\right)=-\frac{1}{\sqrt{3}}$ and the area of the surface (disk with radius $2$) is $4\pi\,.$ Therefore, both sides of (1) should give $-\frac{4\pi}{\sqrt{3}}\,.$
To do all calculations required in (1) explicitly I suggest a simpler example wich contains all essentials:
$$
F=\begin{pmatrix}y\\-x\\0\end{pmatrix}\,,\quad S=\{z=0\}\,,\quad C=\{x^2+y^2=4\}\,.
$$
curl$\,F\cdot\mathbf{n}=\left(\begin{smallmatrix}0\\0\\-2\end{smallmatrix}\right)\cdot\left(\begin{smallmatrix}0\\0\\1\end{smallmatrix}\right)=-2$ so that both sides of (1) should be $-8\pi\,.$ The LHS is $$\int_0^{2\pi}F\cdot\left(\begin{smallmatrix}-2\sin\varphi\\2\cos\varphi\\0\end{smallmatrix}\right)\,d\varphi=\int_0^{2\pi}-4(\sin^2\varphi+\cos^2\varphi)\,d\varphi=-8\pi\,.$$ Using the parametrization $[0,2]\times[0,2\pi)\ni(r,\varphi)\mapsto\left(\begin{smallmatrix}r\cos\varphi\\r\sin\varphi\\0\end{smallmatrix}\right)$ the RHS is seen to be$$\int_0^2\int_0^{2\pi}-2\,\color{red}{r}\,dr\,d\varphi=-8\pi\,.$$ The highlighed $\color{red}{r}$ in that integral is from the unnormalized $(s_u\times s_v)$ in that particular parametrization.
