Positioning a body in space using intersection criteria... (Sorry there is an error in my images below - the coordinates of point A1 should be 0,0,-86.6. I'm not sure how to replace the images. Sorry for the confusion). 
I hope this post is appropriate for this forum.  If not I apologize.  I’m scratching my head over a math problem that I’d like to solve and could use a bit of assistance.  Unfortunately I’m an engineer and while the math is of great interest I don’t even know the realm to start in. I’m sort of assuming this is some sort of matrix problem but I’m not sure. 
The problem involves positioning objects in space to determine a resultant location that satisfies some intersection criteria. Because it’s a bit hard to describe I have a couple of pages describing the basic problem that follow. 
I used a CAD system to graphically solve for one case. I’d ideally like to reduce this to a generalized case that I can use to solve for a given set of inputs.  
I believe that there are two possible solutions (one with the prism above and one with the prism below) however there may be situations with multiple additional solutions that might render this more complicated. If it helps, I can constrain the solution so that the focal point is always a negative X and positive Y value.  I believe this reduces this down to a single solution (but I’m not positive)
I would love not just a solution (which would be cool) but also to learn how this sort of problem is solved). I believe this is a fairly common imaging problem and my CAD system can solve it graphically (or perhaps mathematically behind the scenes) so there must be a solution out there someplace. Thanks in advance for any guidance! 



Starting from @Aretino's answer;
To get rid of big numbers, define $A=\frac a {100}$, $B=\frac b {100}$, $C=\frac c {100}$, write the equations as
$$A^2+B^2-k_1 A B=1 \tag 1$$
$$B^2+C^2-k_2 B C=1 \tag 2$$
$$C^2+A^2-k_3 C A=1 \tag 3$$
$$(1) \implies B_\pm=\frac{1}{2} \left(A k_1\pm\sqrt{A^2 k_1^2-4 A^2+4}\right)$$
$$(3) \implies C_\pm=\frac{1}{2} \left(A k_3\pm\sqrt{A^2 k_3^2-4 A^2+4}\right)$$ So, $(2)$ is left with four combinations but it is just a function of $A$ which is not a big deal (except that it corresponds to an octic polynomial in $A$).
However, Newton method would converge quite fast. For example, for the combination $(B_+,C_+)$, by inspection the root is around $3$ and the iterates would be
$$\left(
\begin{array}{cc}
n & A_n \\
 0 & 3.00000 \\
 1 & 3.80093 \\
 2 & 3.70798 \\
 3 & 3.70520
\end{array}
\right)$$
This leads to
