 Solution

The Stokes theorem states that for an oriented smooth surface $S$ with compatible oriented boundary $C$ and a continuously differentiable vector field,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 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 becomesTherefore the complete endpoint-dependent answer iswhich 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.