Say we have four linear equations in three variables $f_i(x,y,z)$ at $i\in\{1,2,3,4\}$ forming a rank $3$ system.
Is it possible if we eliminate $z$ using $1$ and $2$ to get equation $g_{12}$ and $z$ using $3$ and $4$ to get equation $g_{34}$ the system degenerates to two parallel lines on the plane but by eliminating $z$ using equation $1$ and $3$ to get equation $g_{13}$ and eliminating $z$ using equations $2$ and $4$ to get equation $g_{24}$ we get two non-degenerate equations which provide unique $(x,y)$ solution and by eliminating $z$ using equation $1$ and $4$ to get equation $g_{14}$ and eliminating $z$ using equations $2$ and $3$ to get equation $g_{23}$ we get two non-degenerate equations which provide second unique $(x',y')$ solution?
If not is it always the case $(x=x')$ and $(y=y')$ holds if it happens?
Your system is given by
$ A x = b $
where $A $ is $ 4 \times 3 $, $x$ is a $3 \times 1$ vector, and $b$ is a $4 \times 1$ vector.
To solve this system, the standard way is to form the augmented matrix $[A | b ]$, and reduced it to reduced echelon form. The system will have a unique solution if the reduced row echelon form the augmented matrix is of the form
$\begin{bmatrix} 1 && 0 && 0 && c_1 \\ 0 && 1 && 0 && c_2 \\ 0 && 0 && 1 && c_3 \\ 0 && 0 && 0 && 0 \end{bmatrix}$
If the last entry of the last row is not zero, then there's no solution.
Also, the system can have infinite solutions, for example the 3rd and 4th equations are linearly dependent on the first and second equations. For example, suppose, the augmented matrix is
$\begin{bmatrix} 1 && -2 && 3 && 4 \\ 3 && -1 && 2 && 5 \\ 4 && -3 && 5 && 9 \\ 2 && 1 && -1 && 1 \\ \end{bmatrix}$
One can easily verify that the third row is just the sum of the first two, and that the fourth row is the second row minus the the first. Hence, when reduce the augmented matrix, we'll have two whole rows of zeros. The actual reduced row echelon form is:
$\begin{bmatrix} 1 && 0 && 1/5 && 6/5 \\ 0 && 1 && -7/5 && -7/5 \\ 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 \\ \end{bmatrix}$
Hence, by letting $z = t$ , where $t \in \mathbb{R}$, then
$x = 6/5 - t / 5 $ $y = -7/5 + 7/5 t $
Hence the general solution is
$(x,y,z) = ( 6/5 - t/5, -7/5 + 7/5 t , t )$
One can verify that this satisfies the first and second equations (and thus the third and fourth):
$(6/5 - t/5) - 2 (-7/5 + 7/5 t ) + 3 t = 4$
and
$3 (6/5 - t/5) - (-7/5 + 7/5 t ) +2 t = 5 $