Galois Group of $\mathbb{Q}(\sqrt{3},i\sqrt{3},\zeta_{13})/ \mathbb{Q}$

I would like find the Galois Group of $$\mathbb{Q}(\sqrt{3},i\sqrt{3},\zeta_{13})/ \mathbb{Q},$$

that field extension is Galois because is the splitting field of separable polynomial $$(x^3-3)(x^{13}-1)$$. We know $$[\mathbb{Q}(\sqrt{3},i\sqrt{3}):\mathbb{Q}]=6$$ and $$[\mathbb{Q}(\zeta_{13}):\mathbb{Q}]=12$$, so $$[\mathbb{Q}(\sqrt{3},i\sqrt{3},\zeta_{13}): \mathbb{Q}]\leq 72.$$

Note that $$\text{Gal}(\mathbb{Q}(\sqrt{3},i\sqrt{3})/\mathbb{Q}) \cong S_3$$ is non abelian, so that, implies that $$\sqrt{3} \notin \mathbb{Q}(\zeta_{13})$$, because $$\mathbb{Q}(\zeta_{13})$$ is an abelian Galois field extension (isomorphic to $$\mathbb{Z}_{13}^{*})$$. We can show that $$i\sqrt{3} \notin \mathbb{Q}(\zeta_{13})$$ because the only quadratic extensión of $$\mathbb{Q}(\zeta_{13})$$ is a real extension. ¿which is the degree of the galois extension?

Note that $$\mathbb{Q}(\sqrt{-3},\zeta_{13})=\mathbb{Q}(\zeta_3,\zeta_{13})=\mathbb{Q}(\zeta_{39})$$ and the only degree $$3$$ extension of $$\mathbb{Q}$$ comes from $$\mathbb{Q}(\zeta_{13})$$, so the same argument applies to give you $$[\mathbb{Q}(\zeta_{39})(\sqrt3):\mathbb{Q}]=72$$.
The Galois group is $$S_3\times C_{12}$$.