Essentially I want to vary the action
$$ S_M = \int d^3x \sqrt{-g} \left[- \frac{1}{4} F^{\mu \nu} F_{\mu \nu} - \frac{\alpha}{2} \epsilon^{\mu \nu \rho} A_\mu F_{\nu \rho} \right] $$
with respect to $A_\mu$ in order to find the equation of motion for $A_\mu$. Here, $\epsilon^{\mu \nu \rho}$ is the Levi-Civita tensor. I am trying to follow along with a someone else's solution, which I am pretty certain is correct, but I can't quite understand what they've done.
First, they put the action in terms of differential forms to get
$$ S_M = \int d^3x \left[- \frac{1}{4} F \wedge * F - \frac{\alpha}{2} A \wedge F \right] $$
where $*$ is the Hodge star operator. Here I already have a problem. When I attempt to find the action in terms of differential forms, I get the second term to have a coefficient of just $\alpha$. I got this through the definition of the wedge product given in Carroll
$$ A \wedge F = \frac{6}{2} A_{[\mu} F_{\nu \rho]} = \frac{1}{2\sqrt{-g}} \epsilon^{\mu \nu \rho} A_\mu F_{\nu \rho}. $$
After this, they vary the action in A. Using the fact that $F = dA$, they write out the steps
$ \delta S_M = \int \left[ - \frac{1}{2} d \delta A \wedge * F - \frac{\alpha}{2} \delta A \wedge F + \frac{\alpha}{2} F \wedge \delta A \right] = \int \left[\frac{1}{2} \delta A \wedge d * F - \frac{\alpha}{2} \delta A \wedge F \right] $
From here they set the variation equal to zero to get
$ d * F = \alpha F \hspace{0.5cm} \Rightarrow \hspace{0.5cm} \nabla_\mu F^{\mu \nu} = \alpha \epsilon^{\nu \alpha \beta} F_{\alpha \beta} $
In the variation, I'm not sure how they get these terms. I can't simplify the variation past
$ \delta S_M = \int \left[- \frac{1}{4} d(\delta A) \wedge * F - \frac{1}{4} F \wedge \delta (* F) - \frac{\alpha}{2} \delta A \wedge F + \frac{\alpha}{2} A \wedge d (\delta A) \right] $
Any explanation on how they are getting this would be much appreciated.
