For a nonzero prime ideal $\mathfrak{p}$ of $\mathbb{Z}[\sqrt{-6}]$ which does not divide $2$, does $\mathfrak{p}$ decompose completely in the extension $\mathbb{Q}(\zeta_{24})/\mathbb{Q}(\sqrt{-6})$ of degree $3$ if and only if $\mathfrak{p} = (\alpha)$ for some $\alpha \in \mathbb{Z}[\sqrt{-6}]$ such that $a \equiv 1 \text{ mod }2\mathbb{Z}[\sqrt{-6}]$?

I suspect this is a true based off working out the following example. $5 + 6\sqrt{-6}$ is a prime divisor of the prime number $241$ in $\mathbb{Z}[\sqrt{-6}]$ and $5 + 6\sqrt{-6} \equiv 1 \text{ mod }2\mathbb{Z}[\sqrt{-6}]$. We have$$241 = \prod_{a = 1, 5, 7, , 11, 13, 17, 19, 23} (2 - \zeta_{24}^a), \quad 5 + 6\sqrt{-6} = -\prod_{a = 1, 5, 7, 11}(2 - \zeta_{24}^a).$$