Hexagonal Pyramids.

Let $\lambda$ be the measure of the angle made between two faces.

$\vec{BA} = r\vec{i} + 0\vec{j} + 0\vec{k}$

$\vec{BC} = -\dfrac{r}{2}\vec{i} - \dfrac{\sqrt{3}}{2}r\vec{j} + 0\vec{k}$

$\vec{BQ} = \dfrac{r}{2}\vec{i} - \dfrac{\sqrt{3}}{2}r\vec{j} + h\vec{k}$

$\vec{u} = \vec{BQ} \:\ X \:\ \vec{BA} = 0\vec{i} + hr\vec{j} + \dfrac{\sqrt{3}}{2}r^2\vec{k}$

$\vec{v} = \vec{BQ} \:\ X \:\ \vec{BC} = \dfrac{\sqrt{3}}{2}rh\vec{i} - \dfrac{1}{2}rh\vec{j} - \dfrac{\sqrt{3}}{2}r^2\vec{k}$

$\vec{u} \cdot \vec{v} = -\dfrac{r^2(2h^2 + 3r^2)}{4}$ and $|\vec{u}| = \dfrac{r\sqrt{4h^2 + 3r^2}}{2} = |\vec{v}| \implies \cos(\lambda) = -\dfrac{2h^2 + 3r^2}{4h^2 + 3r^2}$

The lateral surface area $S = \dfrac{3}{2}r\sqrt{4h^2 + 3r^2}$

The volume $V = \dfrac{1}{4\sqrt{3}}r^2h = k \implies h = \dfrac{4\sqrt{3}k}{r^2} \implies$ $S(r) = \dfrac{3}{2r}\sqrt{192k^2 + 3r^6} \implies$

$\dfrac{dS}{dr} = \dfrac{9(r^6 - 32k^2)}{r^2\sqrt{192k^2 + 3r^6}} = 0 \implies r = (4\sqrt{2}k)^{\frac{1}{3}} \implies$ $h = \dfrac{4\sqrt{3}k}{(4\sqrt{2})^{\frac{2}{3}}}$

$\implies \cos(\lambda) = -\dfrac{2}{3} \implies \lambda \approx \boxed{131.81031}$ .