How can I determine the position of the apex of an irregular tetrahedron I have an irregular tetrahedron the base of which is an equilateral triangle.  Knowing the lengths of all sides I need to then determine the position of the apex.
Is there anyone that can offer a bit of guidance on this.

Doing this in $\mathbb{R}^3$, let the three points of the equilateral triangular base of the tetrahedron be $(0,0,0)$, $(0,2\ell,0)$, and $(\sqrt{3}\ell,a,0)$. Suppose the lengths of the edges of the tetrahedron between each of these point and our apex are $a$, $b$, and $c$ respectively. The apex (if it exists) will be the intersection of the three spheres given by the following equations:
$$\begin{cases}
  x^2 + y^2 + z^2 = a^2 \\
  x^2 + (y-2\ell)^2 + z^2 = b^2 \\
  (x-\sqrt{3}\ell)^2 + (y-\ell)^2 + z^2 = c^2 \\
\end{cases}$$
After some rough scratch work, I ended up getting our apex as the point
$$\left(
  \frac{a^2+b^2-2c^2+4\ell^2}{4\sqrt{3}\ell},\quad
  \frac{a^2-b^2+4\ell^2}{4\ell}, \quad
  \sqrt{a^2-x^2-y^2}
\right)$$
The $z$-coordinate is kinda gross, so I left it here in terms of $x$ and $y$.
