Pyramid with a Quadrilateral Base.

Using the law of sines $\implies \dfrac{\sin(2\beta)}{\sin(\beta)} = \dfrac{m}{m - 2} \implies \cos(\beta) = \dfrac{m}{2(m - 2)}$

Using the law of cosines with included $\angle{BAC} \implies m^2 - 4m + 4 = m^2 - 2m + 1 + m^2 - \dfrac{m^2(m - 1)}{m - 2} \implies m^2 - 7m + 6 = 0 \implies$ $(m - 6)(m - 1) = 0 \:\ m \neq 1 \implies m = 6 \implies \overline{AC} = 6a, \:\ \overline{AB} = 5a$ and $\overline{BC} = 4a$

$\cos(\beta) = \dfrac{3}{4} \implies \overline{BD} = \dfrac{5a\sqrt{7}}{4}a$ and $AD = \dfrac{15a}{4}$

Using the above quadrilateral with the given coordinates:

$m_{M_{1}M_{2}} = -\dfrac{5\sqrt{7}}{12} \implies y = -\dfrac{5\sqrt{7}}{12}x + \dfrac{45\sqrt{7a}}{32}$

and,

$m_{M_{3}M_{4}} = \dfrac{5\sqrt{7}}{12} \implies y = \dfrac{5\sqrt{7}}{12}x - \dfrac{45\sqrt{7a}}{32}$

Solving the two equations above we obtain: $x = \dfrac{27a}{8}$ and $y = 0 \implies P(\dfrac{27a}{8},0)$ which is a midpoint of each congruent bimedian.

For triangular face $QDC$ :

$\overline{PM_{2}} = \dfrac{\sqrt{319}a}{8}$ and $h = \overline{PQ} \implies \overline{QM_{2}} =$ $\dfrac{\sqrt{64h^2 + 319a^2}}{8}$ and $\overline{DC} = 4a \implies A = A_{QDC} = \dfrac{1}{2}(\overline{DC})(\overline{QM_{2}}) = \dfrac{a}{4}\sqrt{64h^2 + 319a^2}$

$A_{\triangle{ABC}} = A_{\triangle{ADC}} = \dfrac{15\sqrt{7}a^2}{4} \implies A_{ABCD} = \dfrac{15\sqrt{7}a^2}{2} \implies$ the volume of the given pyramid is $V = \dfrac{5\sqrt{7}a^2}{2} a^2 h = k \implies h = \dfrac{2k}{5\sqrt{7}a^2} \implies A(a) = \dfrac{\sqrt{256k^2 + 55825a^6}}{20\sqrt{7}a^2} \implies$

$\dfrac{dA}{da} = \dfrac{55825a^6 - 128k^2}{10\sqrt{7}a^2\sqrt{256k^2 + 55825a^6}} = 0$

$\implies a = (\dfrac{8\sqrt{2}k}{5\sqrt{2233}})^{\frac{1}{3}} \implies h = \dfrac{2k}{5\sqrt{7}}(\dfrac{5\sqrt{2233}}{8\sqrt{2}k})^{\frac{2}{3}}$

Let $\theta = m\angle{QM_{2}P} \implies \tan(\theta) = \dfrac{8h}{\sqrt{319}a}$

$\dfrac{h}{a} = \dfrac{\sqrt{2}\sqrt{319}}{8} \implies \tan(\theta) = \sqrt{2} \implies \boxed{\theta \approx 54.7356103^\circ}$

$\overline{PD} = \dfrac{\sqrt{709}}{8}$

Let $\lambda = m\angle{QDP} \implies \tan(\lambda) = \dfrac{8}{\sqrt{709}}\dfrac{h}{a} = \sqrt{\dfrac{638}{709}} \implies \boxed{\lambda \approx 43.4892775^\circ}$

$\overline{PC} = \dfrac{21}{8}$

$\omega = m\angle{QPC} \implies \tan(\omega) = \dfrac{8}{21}\dfrac{h}{a} = \dfrac{\sqrt{638}}{21} \implies \boxed{\omega \approx 50.2599337^\circ}$

$\implies \theta + \lambda + \omega = \boxed{148.4848215}$ .