= Solution
The <Stokes theorem> states that for an oriented smooth <surface> $S$ with compatible oriented boundary $C$ and a continuously differentiable <vector field>,
$$
\oint_C F\cdot d\ell=\int_S(\nabla\times F)\cdot n\,dS.
$$
Here $\nabla\times F=(-2,-2,-2)$, whereas a <normal vector> to the specified <plane> is $(1,0,-1)$. Their <dot product> is zero. Thus \b[every closed curve in that <plane> has zero circulation] by the <Stokes theorem>.
The source says “any curve” and displays an ordinary <line integral>, without specifying that it is closed or giving endpoints. An open <curve> need not give zero. On the <plane>, put $z=x-1$ and $dz=dx$; the pulled-back <differential one-form> becomes
$$
F\cdot d\ell=(y-x+1)dx-dy+(x-y)dx=dx-dy=d(x-y).
$$
Therefore the complete endpoint-dependent answer is
$$
\boxed{\int_{P}^{Q}F\cdot d\ell=(x_Q-y_Q)-(x_P-y_P),}
$$
which specializes to zero for a closed <curve>. For example, the path from $(1,0,0)$ to $(2,0,1)$ lies in the <plane> and has integral $1$. This illustrates a <conservative restriction of a vector field to a plane> despite nonzero ambient <curl>.
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