Tetrahedron Craze!

For triangular face $QAC:$

$A = A_{QAC} = \dfrac{a}{2}\overline{QT} = \dfrac{a}{2}\sqrt{h^2 + a^2 r^2}$

$V = \dfrac{1}{4\sqrt{3}} a^2 h^2 = k \implies A(a) = \dfrac{1}{2}\dfrac{\sqrt{48k^2 + a^6 r^2}}{a} \implies$ $\dfrac{dA}{da} = \dfrac{1}{2}\dfrac{2r^2 a^6 - 48 k^2}{a^2\sqrt{48k^2 + a^2 r^2}}$
$a \neq 0 \implies a = (\dfrac{\sqrt{24}k}{r})^{\frac{1}{3}} \implies h = 2\sqrt{3}(\dfrac{kr^2}{3})^{\frac{1}{3}}$

$\implies \tan(\theta) = \dfrac{h}{ar} =\sqrt{2} \implies \boxed{\theta \approx 54.73561^\circ}$

and

$\tan(\lambda) = \dfrac{\sqrt{a^2 r^2 + h^2}}{ar} =$ $\dfrac{\sqrt{48k^2 + a^6r^2}}{a^3 r} = \sqrt{3}$ $\implies \boxed{\lambda = 60^\circ}$

For triangular face $QAB:$

$\vec{u} = ar\vec{i} + ar\vec{j} + h\vec{k}$

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

$\vec{u} X \vec{v} = -\dfrac{\sqrt{3}}{2}ah\vec{i} + \dfrac{ah}{2}\vec{j} + \dfrac{\sqrt{3} - 1}{2}a^2r\vec{k}$

$\implies |\vec{u} X \vec{v}| = \dfrac{a}{2}\sqrt{4h^2 + (4 - 2\sqrt{3})r^2 a^2}, \:\ |\vec{u}| = \sqrt{2r^2a^2 + h^2}$ and $|\vec{v}| = a$

$\implies \sin(\gamma) = \dfrac{\sqrt{4h^2 + (4 - 2\sqrt{3})r^2 a^2}}{2\sqrt{2r^2 a^2 + h^2}} =$ $\sqrt{\dfrac{6 - \sqrt{3}}{8}} \implies \boxed{\gamma \approx 46.92048^\circ}$

$\implies \theta + \lambda + \gamma = \boxed{161.65609^\circ}$ .