= Solution
{c}
Start with the formally gauge-redundant <path integral> $Z=\operatorname{vol}(\mathcal G)^{-1}\int\mathcal DA\,e^{iS[A]}$. Use $A_\mu^\theta=A_\mu+\partial_\mu\theta$, and the generalized <Lorenz gauge condition> $F_\omega[A]=\partial_\mu A^\mu-\omega$. In a perturbative neighborhood where residual zero modes are removed by <boundary conditions>, insert the <Faddeev-Popov gauge-orbit identity>
$$
1=\Delta_{\rm FP}[A]\int\mathcal D\theta\,
\delta[F_\omega(A^\theta)],\qquad
\frac{\delta F_\omega(A^\theta)(x)}{\delta\theta(y)}
=\Box_x\delta^{(4)}(x-y).
$$
Thus $\Delta_{\rm FP}=\det\Box$. <Gauge invariance> lets the $\theta$ integration factor out and cancel $\operatorname{vol}(\mathcal G)$. Average over $\omega$ with the <Gaussian integral> weight $\exp[-i\int\omega^2/(2\xi)]$; the delta functional fixes $\omega=\partial\cdot A$ and gives
$$
\mathcal L_{\rm gf}=-\frac1{2\xi}(\partial\cdot A)^2.
$$
Represent the <functional determinant> by the independent <Grassmann fields> $\bar c,c$. A field-independent determinant sign can be absorbed in normalization, so a convenient convention is
$$
\det\Box\ \propto\ \int\mathcal D\bar c\mathcal Dc\,
\exp\left[-i\int d^4x\,\bar c\Box c\right].
$$
Therefore
$$
\boxed{Z=\mathcal N\int\mathcal DA\mathcal D\bar c\mathcal Dc\,
\exp\left\{i\int d^4x\left[-\frac14F_{\mu\nu}F^{\mu\nu}
-\frac1{2\xi}(\partial\cdot A)^2-\bar c\Box c\right]\right\}.}
$$
The <Faddeev-Popov ghosts> are free and decouple because $\Box$ is independent of $A$. For a <Non-Abelian gauge theory>, an infinitesimal <gauge transformation> gives $\delta A_\mu^a=D_\mu^{ab}\theta^b$, so the <Faddeev-Popov operator> is $\partial^\mu D_\mu^{ab}[A]$ and
$$
\mathcal L_{\rm gh}=-\bar c^a\partial^\mu D_\mu^{ab}[A]c^b.
$$
It contains a <ghost-gluon vertex>: these <Faddeev-Popov ghosts> do not decouple in covariant <gauge fixing> of <Lorenz gauge> type.
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