Coordinate Challenge

Suppose the intersection points with the $xy$ plane are $\vec{I_1},\vec{I_2},\vec{I_3}$ , and that the submerged vertex is $\vec{P}$ . The three vectors from $\vec{P}$ to $\vec{I_1},\vec{I_2},\vec{I_3}$ must all be perpendicular.

$(I_{1x} - P_x)(I_{2x} - P_x) + (I_{1y} - P_y)(I_{2y} - P_y) + (I_{1z} - P_z)(I_{2z} - P_z) = 0 \\ (I_{2x} - P_x)(I_{3x} - P_x) + (I_{2y} - P_y)(I_{3y} - P_y) + (I_{2z} - P_z)(I_{3z} - P_z) = 0 \\ (I_{3x} - P_x)(I_{1x} - P_x) + (I_{3y} - P_y)(I_{1y} - P_y) + (I_{3z} - P_z)(I_{1z} - P_z) = 0$

Solving this non-linear system and choosing the negative-z result gives $(P_x,P_y,P_z) \approx (-3.6361,5.5186,-6.7865)$ . It is then a fairly trivial matter to reconstruct the eight cube vertices using the side lengths and basis vectors consisting of unitized versions of $(\vec{I_1} - \vec{P})$ , $(\vec{I_2} - \vec{P})$ , $(\vec{I_3} - \vec{P})$ . The highest vertex ends up being $\approx (1.3235,-5.3906,25.7156)$ , and the sum of its coordinates is approximately $21.6485$ .