First, see my comment, following your query.
Since you are trying, not only to solve the equations, but also to express
$b$ and $c$ in terms of $a$, I would have approached it differently.
$$2a + 3b - 4c = 7. \tag1$$
$$a - b + 2c = 6. \tag2$$
Multiplying equation (2) above by $(3)$ gives
$$3a - 3b + 6c = 18.\tag3$$
Adding equations (3) and (1):
$$5a + 2c = 25 \implies c = \frac{25 - 5a}{2}. \tag4$$
Now, repeat the process to express $b$ in terms of $a$.
Multiplying equation (2) by $(2)$ gives
$$2a - 2b + 4c = 12. \tag5$$
Adding equations (5) and (1):
$$4a + b = 19 \implies b = 19 - 4a. \tag6$$
At this point, you are actually done. $(a)$ can be chosen to be any number.
Once $(a)$ is chosen, $(b)$ and $(c)$ are computed via equations (4) and (6) above.
Then, you have generated a solution $(a,b,c).$
Set $x_5 = c $ and $x_6 = c^2 $, then the system becomes
$\begin{bmatrix} 2&&1&&5&&4&&0&&-1 \\ 1 && -1&& 1 &&-1&&-4&&0 \\ 1 && 0 && 2 && 1 && -1 && 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \\ 3 \end{bmatrix} $
Solving using Gauss-Jordan elimination, the solutions are
$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{bmatrix} = \begin{bmatrix} 5 \\ -8\\ 0 \\ 0 \\ 2 \\ 0 \end{bmatrix} + t_1 \begin{bmatrix} -2 \\ -1 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + t_2 \begin{bmatrix} -1 \\ - 2 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t_3 \begin{bmatrix} -1 \\ 3 \\ 0 \\ 0 \\ -1 \\ 1 \end{bmatrix} $
From which it follows that
$x_5 = c = 2 - t_3 $ and $ x_6 = c^2 = t_3 $
Hence, we must have $2 - c = c^2 $, i.e. $c^2 + c - 2 = 0$ which can be factored as $ (c + 2)(c - 1) = 0 $. Therefore, the two values of $c$ for which there is a solution are $c = -2 $ and $c = 1$.
For $c = -2$, we get $t_3 = 4 $ and for $c = 1$ we get $t_3 = 1 $
Therefore, there are two corresponding sets of solutions:
Case I: for $c = -2$ we get,
$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \\ 0 \\ 0 \end{bmatrix} + t_1 \begin{bmatrix} -2 \\ -1 \\ 1 \\ 0 \end{bmatrix} + t_2 \begin{bmatrix} -1 \\ - 2 \\ 0 \\ 1 \end{bmatrix} $
Case II: for $c = 1$, we get,
$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 4 \\ -5 \\ 0 \\ 0 \end{bmatrix} + t_1 \begin{bmatrix} -2 \\ -1 \\ 1 \\ 0 \end{bmatrix} + t_2 \begin{bmatrix} -1 \\ - 2 \\ 0 \\ 1 \end{bmatrix} $
Best Answer
Note that\begin{align}1&=14-13\\&=rs+t-r-st\\&=s(r-t)-(r-t)\\&=(s-1)(r-t).\end{align}Since we are dealing with integers here, there are not many possibilities. You will have to have $s=2$ and $r=t+1$ or $s=0$ and $r=t-1$. Can you take it from here?