= Solution
{c}
Expand the <gauge fixing> square:
$$
\boxed{\mathcal L_{\mathrm{gf}}=-\frac1{2\xi}(\partial\cdot A)^2-\frac\lambda\xi(\partial\cdot A)A_\mu A^\mu-\frac{\lambda^2}{2\xi}(A_\mu A^\mu)^2.}
$$
The second term is a derivative cubic <photon> interaction; the third is a quartic <photon> <contact interaction>. Both arise from <nonlinear Abelian gauge fixing>, rather than an additional physical electromagnetic force.
For $\xi=1$, differentiating the quartic term with respect to four identical <gauge fields> gives eight copies for each pairing, hence
$$
V_4^{\alpha\beta\gamma\delta}=-4i\lambda^2(g^{\alpha\beta}g^{\gamma\delta}+g^{\alpha\gamma}g^{\beta\delta}+g^{\alpha\delta}g^{\beta\gamma}).
$$
With the Fourier convention $A(x)=\int d^4p\,(2\pi)^{-4}e^{ipx}A(p)$ and all <four-momenta> incoming, the differentiated cubic term gives
$$
V_3^{\alpha\beta\gamma}=2\lambda(g^{\beta\gamma}p_1^\alpha+g^{\alpha\gamma}p_2^\beta+g^{\alpha\beta}p_3^\gamma),
$$
matching the supplied <Feynman rules>. Reversing the Fourier convention reverses this cubic rule's sign; each exchange graph contains two such <Feynman vertices>, so its contribution is unchanged.
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