System of Six Equations, Line Conic Degeneration and Decomposition If we have the line conics
\begin{align}
\Sigma & \equiv \mathscr{A} l^2 + \mathscr{B} m^2 + \mathscr{C} n^2 + 2 \mathscr{F} m n + 2 \mathscr{G} n l + 2 \mathscr{H} l m = 0 \\
\Sigma' & \equiv \mathscr{A}' l^2 + \mathscr{B}' m^2 + \mathscr{C}' n^2 + 2 \mathscr{F}' m n + 2 \mathscr{G}' n l + 2 \mathscr{H}' l m = 0
\end{align}
then we can choose $\lambda$ such that the pencil of line conics $\Sigma + \lambda \Sigma' = 0$ is degenerate, it will factor into a point-pair: $$(A l + B m + C n) (A' l + B' m + C' n) = 0$$ If we expand this equation, we get $$A A' l^2 + B B' m^2 + C C' n^2 + (B C' + C B') m n + (A C' + C A') n l + (A B' + B A') l m = 0$$ Then we see that the six coefficients $A, B, C, A', B', C'$ are given by the system of six equations\begin{align}
 A A' & = \mathscr{A} + \lambda \mathscr{A}' \\
 B B' & = \mathscr{B} + \lambda \mathscr{B}' \\
 C C' & = \mathscr{C} + \lambda \mathscr{C}' \\
 B C' + C B' & = 2 (\mathscr{F} + \lambda \mathscr{F}') \\
 A C' + C A' & = 2 (\mathscr{G} + \lambda \mathscr{G}') \\
 A B' + B A' & = 2 (\mathscr{H} + \lambda \mathscr{H}')
\end{align}
How do I go about solving this system for $A, B, C, A', B', C'$? None of my owned calculators can do it. They run out of resources. I tried solving it by hand but after writing multiple pages I am afraid I will make a mistake and it will all go to waste. If it is not possible to gift me the solutions, is there any easy way to solve this system?
A range of conics has three point-pairs, so there should be three different solutions to the equation.

\begin{align*}
 A A' & = \mathscr{A} \tag1\\
 B B' & = \mathscr{B}\tag2\\
 C C' & = \mathscr{C}\tag3 \\
 B C' + C B' & = 2 \mathscr{F}\tag4 \\
 A C' + C A' & = 2 \mathscr{G} \tag5\\
 A B' + B A' & = 2 \mathscr{H}\tag6
\end{align*}
From $(1)(2)(3)$,
$$A'=\frac{\mathscr{A}}{A},\quad B'=\frac{\mathscr B}{B},\quad C'=\frac{\mathscr C}{C}$$
Substituting these into $(5)(6)$ gives
$$(5)\implies \frac{C^2}{A^2}\mathscr{A}  - 2\mathscr{G}\frac CA +\mathscr C=0\implies \frac CA=\frac{\mathscr G\pm\sqrt{{\mathscr G}^2-\mathscr A\mathscr C}}{\mathscr A}:=p$$
$$(6)\implies  \frac{B^2}{A^2}\mathscr{A} -2\frac BA \mathscr{H}+\mathscr B =0\implies \frac BA=\frac{\mathscr H\pm\sqrt{{\mathscr H}^2-\mathscr A\mathscr B}}{\mathscr A}:=q$$
giving $$B=qA,\quad C=pA$$
Substituting these into $(4)$ gives
$$q\frac{\mathscr C}{p}+p\frac{\mathscr B}{q}=2\mathscr F\tag7$$
So, as a result, the system has solutions 
$$(A,B,C,A',B',C')=\left(r,qr,pr,\frac{\mathscr A}{r},\frac{\mathscr B}{qr},\frac{\mathscr C}{pr}\right)$$
where $r\not=0$ is a real number and
$$p=\frac{\mathscr G\pm\sqrt{{\mathscr G}^2-\mathscr A\mathscr C}}{\mathscr A},\quad q=\frac{\mathscr H\pm\sqrt{{\mathscr H}^2-\mathscr A\mathscr B}}{\mathscr A}$$ if and only if $(7)$ holds.
