Is the conformal Killing factor always an eigenvalue of the Laplacian?

Question

Suppose $(M, g)$ is a pseudo-Riemannian manifold and $\xi_{\mu}$ is a conformal Killing field, i.e. $$ \nabla_{\nu} \xi_{\mu} + \nabla_{\mu} \xi_{\nu} = \kappa g_{\mu \nu} $$ for some smooth function $\kappa$. \begin{align*} (\text{tr} g-2)\nabla_\mu\nabla_\nu \kappa + g_{\mu\nu}\nabla^\alpha\nabla_\alpha\kappa&= g^{\alpha\beta}\left(g_{\mu\nu}\nabla_\alpha\nabla_\beta+g_{\alpha\beta}\nabla_\mu\nabla_\nu-g_{\alpha\mu}\nabla_\beta\nabla_\nu-g_{\beta\nu}\nabla_\mu\nabla_\alpha\right)\kappa \\ &= g^{\alpha\beta}\left[ \nabla_\alpha\nabla_\beta(\nabla_{\mu} \xi_{\nu} + \nabla_{\nu} \xi_{\mu})+\nabla_\mu\nabla_\nu(\nabla_{\alpha} \xi_{\beta} + \nabla_{\beta} \xi_{\alpha}) \right. \\ &\left. \quad \quad \ -\nabla_\beta\nabla_\nu(\nabla_{\mu} \xi_{\alpha} + \nabla_{\alpha} \xi_{\mu})-\nabla_\mu\nabla_\alpha(\nabla_{\beta} \xi_{\nu} + \nabla_{\nu} \xi_{\beta}) \right] \\ &= g ^{\alpha \beta} \left[ (\nabla _{\alpha \beta \mu} ^3 - \nabla_{\mu \alpha \beta} ^3)\xi_{\nu} + (\nabla_{\alpha \beta \nu} ^3 - \nabla_{\beta \nu \alpha} ^3)\xi_{\mu} \right. \\ &\left. \quad \quad \quad \ (\nabla_{\mu \nu \beta} ^3 - \nabla_{\beta \nu \mu} ^3)\xi_{\alpha} + (\nabla_{\mu \nu \alpha} ^3 - \nabla_{\mu \alpha \nu} ^3)\xi_{\beta}\right] \tag{*} \end{align*} where $\nabla_{\alpha \beta \mu} ^3 = \nabla_{\alpha} \nabla_{\beta} \nabla_{\mu}$. When the Riemann curvature tensor vanishes, the covariant derivatives commute and we obtain: $$ (\text{tr} g -2) \nabla_{\mu} \nabla_{\nu} \kappa + g_{\mu \nu }\Delta_g \kappa = 0. $$ Otherwise, I do not know how to get rid of the dependence on $\xi$ on the right hand side. I have been toying around with the Riemann tensor but I do not get the necessary cancellations. Because there are four terms, I believe that each term must be independent of $\xi$, since otherwise, the only other way I see any cancellations occurring is via Bianchi identities, which have only 3 terms. This is of course just a heuristic but even then, I have not managed to simplify the equation enough to make it depend only on $\kappa$. Is there a way to make the the RHS of $(*)$ independent of $\xi$?