= Solution
Put $R=\sqrt8$ and parametrize the circle counterclockwise, as viewed from positive $z$, by $\mathbf r(\phi)=(R\cos\phi,R\sin\phi,3)$ for $0\leq\phi\leq2\pi$. The restricted field is
$$
V(\phi)=2R^3\cos^3\phi+R^2\sin^2\phi+27-R\sin\phi.
$$
\b[The PDF does not specify the differential in this <line integral>.] An oriented <vector line integral weighted by a scalar field> uses $d\mathbf r=(-R\sin\phi,R\cos\phi,0)\,d\phi$. Its two nonzero components are
$$
\begin{aligned}
\oint_\Gamma V\,dx&=-R\int_0^{2\pi}V(\phi)\sin\phi\,d\phi=\pi R^2,\\
\oint_\Gamma V\,dy&=R\int_0^{2\pi}V(\phi)\cos\phi\,d\phi
=2R^4\int_0^{2\pi}\cos^4\phi\,d\phi=\frac32\pi R^4.
\end{aligned}
$$
All other terms vanish by sine-cosine symmetry. Thus this interpretation gives
$$
\boxed{\oint_\Gamma V\,d\mathbf r=(8\pi,96\pi,0)}.
$$
If the intended convention is the scalar <arc length> <integral>, then $ds=R\,d\phi$ and
$$
\oint_\Gamma V\,ds=R\left(R^2\int_0^{2\pi}\sin^2\phi\,d\phi+54\pi\right)
=\boxed{124\sqrt2\,\pi}.
$$
That scalar <integral> is independent of traversal direction. Giving both values retains the actual distinction left unspecified by the wording.
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