This is an alternative to my other solution. Unlike the other solution, the plane of intersection between the two vision cones is constructed algebraically, without the need for the iterative Newton-Raphson iteration to search for the plane, and without trial and error.
We have two "vision" cones from the two cameras, each containing the $3D$ ellipse on its surface. We want to find the intersection of these two cones.
The equations of the cones are of the form
$ (r - e_1)^T A_1 (r - e_1) = 0 $
and
$ (r - e_2)^T A_2 (r - e_2) = 0 $
where $ r = [x, y, z]^T $ and $e_1, e_2$ are the $3D$ coordinates of the two cameras. If we define $ r = [x, y, z, 1]^T$ (homogenous coordinates) then the above two equations become
$ r^T Q_1 r = 0$
and
$ r^T Q_2 r = 0 $
where
$ Q_1 = \begin{bmatrix} A_1 && - A_1 e_1 \\ -e_1^T A_1 && e_1^T A_1 e_2 \end{bmatrix} $
And $Q_2$ is defined similarly. Now define $Q_\alpha$ as follows
$ Q_\alpha = Q_1 + \alpha Q_2 $
Then $\det(Q_\alpha) $ is a quartic polynomial in $\alpha$. Find its roots, and select one of the roots.
Note that if $r$ satisfies $ r^T Q_1 r = r^T Q_2 r = 0 $ then it satisfies $ r^T Q_\alpha r = 0 $. Therefore, all possible solutions for the intersection of the two quadrics (the two cones) must satisfy $ r^T Q_\alpha r = 0 $.
Diagonalizing $Q_\alpha$ into
$ Q_\alpha = R D R^T $
For this particular application, the diagonal matrix $D$ can always be re-arranged as follows:
$ D = \begin{bmatrix} D_{11} && 0 && 0 && 0 \\
0 && D_{22} && 0 && 0 \\ 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 \end{bmatrix} $
where $D_{11} \gt 0 $ and $ D_{22} \lt 0 $. Define $p = R^T r $, then
$ p^T D p = 0 $
and the general soltion for $p$ is
$ p = t_1 \begin{bmatrix} \dfrac{1}{\sqrt{D_{11}}} \\ \pm \dfrac{1}{\sqrt{-D_{22}}} \\ 0 \\ 0 \end{bmatrix} + t_2 \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} + t_3 \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} $
Since $ r = R p $ has its $4$-th coordinate equal to $1$, then we have to impose
$ [0, 0, 0, 1] R p = 1 $
And this puts a linear constraint on the three parameters $t_1, t_2, t_3$, so that now $p$ is expressable as follows
$ p = p_0 + t_1 p_1 + t_2 p_2 $
From which it follows that
$ r = R p = R p_0 + t_1 R p_1 + t_2 R p_2 $
And this is the parametric equation of the plane(s) of the $3D$ ellipse. Corresponding to the sign choice of $\dfrac{1}{\sqrt{-D_{22}}}$, we have two possible planes.
What remains is to intersect these two planes with either cone to obtain the two equations of the two possible $3D$ ellipses.
This completes the solution of the problem.