= Solution
Away from the origin, the given <vector field> is the <gradient> of the globally single-valued radius:
$$
\mathbf F=\nabla r,\qquad r=\sqrt{x^2+y^2+z^2}.
$$
The sphere has centre $(5,3,2)$, at distance $\sqrt{38}$ from the origin, and radius one. Its intersection circle is therefore wholly away from the origin, since every point on the sphere has distance at least $\sqrt{38}-1>0$. Along the closed curve the <chain rule> gives $\mathbf F\cdot d\mathbf x=dr$. \b[Consequently, independently of orientation,]
$$
\boxed{\oint_C\mathbf F\cdot d\mathbf x=0.}
$$
The intersection is nondegenerate: the centre's distance from the plane is $4/\sqrt{35}<1$. No parametrization of the resulting circle is needed because $r$ is a <potential of a conservative vector field> throughout the punctured space.
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