Consider the following scenario.
The position vectors of the vertices A, B and C , of a tetrahedron are $(1,1,1),(1,0,0), (3,0,0)$ respectively. The altitude from the vertex D to the opposite face $\triangle ABC$ meets the median line through A of the $\triangle ABC$ at point E. The length of side AD is $4$ and volume of the tetrahedron is $\frac{2\sqrt{3}}{3}$.
I am thinking that we get a unique tetrahedron here and thus a unique position for E
But using the fact that E divides AD in ratio say $k:1$ and that $\triangle AED$ is a right triangle. I am getting two values of $k$.
Is one of them an extra solution? If yes, how do I reject the extra one? If no, can you tell how two positions are possible.