Fitting a cone to four points

The vector equation through $\overrightarrow{AB}$ can be given as $(x, y, z) = (0, 0, 30) + (0, 0, -30)t_1$ , through $\overrightarrow{AC}$ as $(x, y, z) = (0, 0, 30) + (20, 0, -30)t_2$ , and through $\overrightarrow{AD}$ as $(x, y, z) = (0, 0, 30) + (10, 10, -30)t_3$ .

Let $E$ be on $\overrightarrow{AC}$ such that $|\overrightarrow{AE}| = |\overrightarrow{AB}|$ , and let $F$ be on $\overrightarrow{AD}$ such that $|\overrightarrow{AF}| = |\overrightarrow{AB}|$ . Since $E$ is on $\overrightarrow{AC}$ and $|\overrightarrow{AE}| = |\overrightarrow{AB}|$ , $(20^2 + 0^2 + (-30)^2)t_2^2 = 0^2 + 0^2 + (-30)^2$ , which solves to $t_2 = \frac{3}{\sqrt{13}}$ , which means that $E$ is at $(x, y, z) = (0, 0, 30) + (20, 0, -30)\frac{3}{\sqrt{13}} = (\frac{60\sqrt{13}}{13}, 0, \frac{390 - 90\sqrt{13}}{13})$ . Similarly, since $F$ is on $\overrightarrow{AD}$ and $|\overrightarrow{AF}| = |\overrightarrow{AB}|$ , $(10^2 + 10^2 + (-30)^2)t_3^2 = 0^2 + 0^2 + (-30)^2$ , which solves to $t_2 = \frac{3}{\sqrt{11}}$ , which means that $F$ is at $(x, y, z) = (0, 0, 30) + (10, 10, -30)\frac{3}{\sqrt{11}} = (\frac{30\sqrt{11}}{11}, \frac{30\sqrt{11}}{11}, \frac{330 - 90\sqrt{11}}{11})$ .

The points $B(0, 0, 0)$ , $E(\frac{60\sqrt{13}}{13}, 0, \frac{390 - 90\sqrt{13}}{13})$ , and $F(\frac{30\sqrt{11}}{11}, \frac{30\sqrt{11}}{11}, \frac{330 - 90\sqrt{11}}{11})$ must be on a plane that cuts the cone perpendicular to its axis, and must be on a circle in which that axis passes through its center $O$ . The equation of the plane through $B$ , $E$ , and $F$ is:

$(3\sqrt{143} - 13\sqrt{11})x + (3\sqrt{143} + 13\sqrt{11} - 22\sqrt{13})y + 2\sqrt{143}z = 0$

the equation of the plane on the perpendicular bisector of $B$ and $E$ is

$2\sqrt{13}x + (13 - 3\sqrt{13})z = 390 - 90\sqrt{13}$

and the equation of the plane on the perpendicular bisector of $B$ and $F$ is

$\sqrt{11}x + \sqrt{11}y + (11 - 3\sqrt{11})z = 30(11 - 3\sqrt{11})$

and these three equations intersect at $O \approx (8.31904, 0.38052, 2.52407)$ .

This makes $\overrightarrow{AO} \approx (8.31904, 0.38052, -27.47593)$ , the unit vector $(a_x, a_y, a_z) \approx \frac{1}{\sqrt{8.31904^2 + 0.38052^2 + (-27.47593)^2}}(8.31904, 0.38052, -27.47593) \approx (0.28976, 0.01325, -0.95701)$ , and the semi-vertical angle $\theta_c \approx \cos^{-1} \Big( \frac{(0, 0, -30) \bullet (8.31904, 0.38052, -27.47593)}{\sqrt{0^2 + 0^2 + (-30)^2}\sqrt{8.31904^2 + 0.38052^2 + (-27.47593)^2}} \Big) \approx 0.29429$ .

Therefore, $a_x + a_y + a_z + \theta_c \approx 0.28976 + 0.01325 + (-0.95701) + 0.29429 \approx \boxed{-0.3597}$ .