Tetrahedron Bonanza

$\vec{u} = -a\vec{i} + a\vec{j} + h\vec{k}$ and $\vec{v} = -2a\vec{i} + \dfrac{2a}{\sqrt{3}}\vec{j} + 0\vec{k}$

$\vec{u} X \vec{v} = \dfrac{2a}{\sqrt{3}}(-h\vec{i} - \sqrt{3}h\vec{j} + (\sqrt{3} - 1)a\vec{k})$

$|\vec{u}| = \sqrt{2a^2 + h^2}$ and $|\vec{u} X \vec{v}| = \dfrac{2a}{\sqrt{3}}\sqrt{4h^2 + (4 - 2\sqrt{3})a^2}$ $\implies d = \dfrac{2a\sqrt{4h^2 + (4 - 2\sqrt{3})a^2}}{\sqrt{3}\sqrt{2a^2 + h^2}} \implies$ The area $A = A_\triangle{QBO} = \dfrac{2a\sqrt{4h^2 + (4 - 2\sqrt{3})a^2}}{\sqrt{3}}$ .

The volume $V = \dfrac{4}{\sqrt{3}} a^2 h = k \implies h = \dfrac{\sqrt{3} k}{4a^2} \implies$ $A(a) = \dfrac{4}{\sqrt{3}}\dfrac{\sqrt{3k^2 + 4(4 - 2\sqrt{3})a^6}}{a} \implies$ $\dfrac{dA}{da} = \dfrac{4(8(4 - \sqrt{3})a^6 - 3k^2}{\sqrt{3} a^2\sqrt{3k^2 + 4(4 - 2\sqrt{3})a^6}} = 0$

$a \neq 0 \implies a = (\dfrac{3 k^2}{16(2 - \sqrt{3})})^{\frac{1}{6}} \implies h = (\dfrac{\sqrt{3}(2 - \sqrt{3})k}{4})^{\frac{1}{3}}$ .

$|\vec{u}| = (\dfrac{\sqrt{3}k}{4\sqrt{2 - \sqrt{3}}})^{\frac{1}{3}}\sqrt{4 - \sqrt{3}}$ , $\:\ |\vec{v}| = \dfrac{4a}{\sqrt{3}} = (\dfrac{16k}{3\sqrt{2 - \sqrt{3}}})^{\frac{1}{3}}$ and $|\vec{u} X \vec{v}| = \dfrac{2\sqrt{12 - 6\sqrt{3}} * k^{\frac{2}{3}}}{16\sqrt{3} (2 - \sqrt{3}))^{\frac{1}{3}}}$

$\implies \sin(\theta) = \dfrac{|\vec{u} X \vec{v}|}{|\vec{u}| |\vec{v}|}$ $= \dfrac{\sqrt{12 - 6\sqrt{3}}}{2\sqrt{4 - \sqrt{3}}} \implies \theta \approx \boxed{24.8960906^\circ}$