I have an ellipse in $x$-$y$ plane with eccentricity $e$, whose semi-major (along $x$-axis) and semi-minor (along $y$-axis) axes lengths are known to be $A$ and $B$. I rotate this ellipse about $y$-axis such that it makes an angle $\epsilon$ with the $x$-$y$ plane. Let's call this new ellipse $E_1$ (shown in red). Further, I rotate $E_1$ by $120^\circ$ anticlockwise about the $z$-axis, and I call this ellipse $E_2$ (shown in blue).
The problem is to find the point on $E_2$ that is at a distance of $L$ from the end-point of the semi-major axis of $E_1$.
The known quantities are: $e$ (eccentricity), the angle $\epsilon$, the coordinates $(x_0,y_0,z_0)$ of the end of semi-major axis of $E_1$, and the distance $L$. The ellipses are shown in the diagram.
Let rotate coordinates around Z first so the blue ellipse get along with the new axes.
$p=-\frac 1 2 x+\frac {\sqrt 3}2 y$
$q=\frac {\sqrt 3} 2 x+\frac 1 2 y$
Now we can easily define ellipsoid around the blue ellipse by squizing sphere $p^2+q^2+z^2=A^2$ along q axis. Its equation is $p^2+(qA/B)^2+z^2=A^2$
This ellipsoid intersects with a plane to make the blue ellipse. Equation of the plane is $z=\frac {z_0}{x_0}p$
Now it's left to find intersection with the sphere $(p+\frac{x_0}{2})^2+(q-\frac{x_0\sqrt3}{2})^2+(z-z_0)^2=L^2$
Good luck with solving this system...