You can check why the range for $t$ I provided in my comment will work by checking values of $t$ within that interval - what do you get when $t=0$? $t=\dfrac{\pi}{2}$? and so on.
The integral becomes
$$\begin{align*}I&=\oint_C(y-z)\,\mathrm{d}x+(z-x)\,\mathrm{d}y+(x-y)\,\mathrm{d}z\\[1ex]
&=\int_0^{2\pi}\bigg((a\sin t-b+b\cos t)(-a\sin t)+(b-b\cos t-a\cos t)(a\cos t)+(a\cos t-a\sin t)(b\sin t)\bigg)\,\mathrm{d}t\\[1ex]
&=\int_0^{2\pi}\bigg(-a^2-ab+ab\cos t+ab\sin t\bigg)\,\mathrm{d}t\end{align*}$$
If you want to use Stokes,
$$
\int_C F\cdot dr=\int\!\!\!\int_S\text{curl} F\cdot dS,
$$
where $S$ is a surface that has $C$ as boundary.
The advantage here is that
$$
\text{curl}\,F=\begin{vmatrix}i&j&k\\ D_x& D_y& D_z\\-y^2&x&z^2\end{vmatrix}
=(0,0,1+2y)
$$
is simple.
If we take $S$ as the disc in the plane $y+z=2$ and bounded by the cylinder, we can parametrize it by
$$
u(r,t)=(r\cos t, r\sin t, 2-r\sin t),\ \ 0\leq r\leq1,\ 0\leq t\leq 2\pi.
$$
Then normal vector is
$$
u_r\times u_t=\begin{vmatrix}i&j&k\\ \cos t& \sin t& -\sin t\\ -r \sin t& r\cos t&-r\cos t\end{vmatrix}
=(0,r,r).
$$
Then
$$
\int_C F\cdot dr=\int\!\!\!\int_S\text{curl} F\cdot dS
=\int_0^1\int_0^{2\pi}(0,0,1+2r\sin t)\cdot(0,r,r)\,dt\,dr\\
=\int_0^1\int_0^{2\pi}(r+2r^2\sin t)\,dt\,dr
=\int_0^12\pi r\,dr=\pi.
$$
In this case, it is also easy to solve the line integral directly: using the parametrization $c(t)=(\cos t, \sin t,2-\sin t)$, $0\leq t\leq 2\pi$, we get
$$
\int_C F\cdot dr=\int_0^{2\pi}(-\sin^2t,\cos t,(2-\sin t)^2)\cdot(-\sin t,\cos t,-\cos t)\,dt=\int_0^{2\pi}(-\sin^3t+\cos^2 t-(2-\sin t)^2\cos t )\,dt=\pi.
$$
Best Answer
Parameterize the surface $S:z=1-x-y$ with the vector function $\vec{r}(x,y)=\left\langle x,y,1-x-y\right\rangle$, where $(x,y)$ belongs to the unit disk $D:x^2+y^2\leq 1$.
Compute the normal vector to $S$ as $\vec{r}_{x}\times\vec{r}_{y}=\langle-z_x,-z_y,1\rangle$.
Calculate $\textrm{curl }\vec{F}$ and plug in $\vec{r}(x,y)$ to get $\textrm{(curl }\vec{F})(\vec{r}(x,y))$.
Take the dot product $\textrm{(curl }\vec{F})(\vec{r}(x,y))\cdot (\vec{r}_{x}\times\vec{r}_{y})$.
Evaluate the surface integral $\iint_S \textrm{curl }\vec{F} \cdot d\vec{S}=\iint_D \textrm{(curl }\vec{F})(\vec{r}(x,y))\cdot (\vec{r}_{x}\times\vec{r}_{y})\,dA$ by converting to polar coordinates.