Fermat point for a tetrahedron

For a $\triangle ABC$ , the Fermat-Torricelli point is the point $P$ that minimizes the sum of distances $\overline{PA} + \overline{PB} + \overline{PC}$ . For a tetrahedron $ABCD$ , we can define a similar point $P$ that minimizes the sum of distances $\overline{PA} + \overline{PB} + \overline{PC} + \overline{PD}$ . Suppose you're given the tetrahedron with vertices $A(10, 0, 0) , B(5, 12, 0) , C(-5, 2, 0) , D(0, 0, 10)$ , numerically find $P(x, y, z)$ that is the minimizing point for the sum of distances from $P$ to the vertices. As your answer, enter $\lfloor 10^4 (x + y + z) \rfloor$