I'm starting to get the hang of this Gauss-Jordan stuff - well, I have never done a system with infinite solutions, so I decided to try this one. You can scroll to the bottom instead to see my doubts though.
$$\begin {cases} -2x+6y-4z=-28 \\ -x+3y-z = -8 \\5x-15y+10z=70\\ x-3y=2\end {cases}$$
So let's reduce this matrix.
\begin{bmatrix} \color{red}{-2} & 6 & -4 & -28 \\ -1 & 3 & -1 & -8 \\ 5 & -15 & 10 & 70 \\ 1 & -3 & 0 & 2 \end{bmatrix}
Gotta make that $\color{red}{-2}$ into a $1$:
$$-\frac{1}{2}f_1 \implies$$
\begin{bmatrix} 1 & -3 & 2 & 14 \\ \color{red}{-1} & 3 & -1 & -8 \\ 5 & -15 & 10 & 70 \\ 1 & -3 & 0 & 2 \end{bmatrix}
Get rid of $\color{red}{-1}$...
$$f_1+f_2$$
\begin{bmatrix} 1 & -3 & 2 & 14 \\ 0 & 0 & 1 & 6 \\ \color{red}{5} & -15 & 10 & 70 \\ 1 & -3 & 0 & 2 \end{bmatrix}
The $\color{red}{5}$...
$$-5f_4 + f_3 \implies$$
\begin{bmatrix} 1 & -3 & 2 & 14 \\ 0 & 0 & 1 & 6 \\ 0 & 0 & 10 & 60 \\ \color{red}{1} & -3 & 0 & 2 \end{bmatrix}
Now for the $\color{red}{1}$:
$$-f_1+f_4\implies$$
\begin{bmatrix} 1 & -3 & 2 & 14 \\ 0 & 0 & 1 & 6 \\ 0 & 0 & \color{red}{10} & \color{red}{60} \\ 0 & 0 & -2 & -12 \end{bmatrix}
Well, for $\color{red}{10}$ and $\color{red}{60}$,
$$5f_4+f_3\implies$$
\begin{bmatrix} 1 & -3 & 2 & 14 \\ 0 & 0 & 1 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \color{red}{-2} & \color{red}{-12} \end{bmatrix}
Then for $\color{red}{-2}$ and $\color{red}{-12}$,
$$2f_2+f_4\implies$$
\begin{bmatrix} 1 & -3 & \color{red}{2} & 14 \\ 0 & 0 & 1 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
Finally, for $\color{red}{2}$,
$$-2f_2+f_1$$
\begin{bmatrix} 1 & -3 & 0 & 2 \\ 0 & 0 & 1 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
And it is both staggered and reduced, I think.
With this, I know that the $range$ of both the matrix and its augmented form is 2 - since both have the same range, then there is a solution.
Next, since $2 < 3$ (the number of columns), I can conclude that the system has an infinite number of solutions, right? They depend on $3 - 2 = 1$ parameter.
The solutions are:
$$\begin{cases} x = 2 + 3y \\ y = ? \\ z = 6\end{cases}$$
We assign $y$ some $t \in \mathbb{R}$ and conclude that the solution set is
$$S = \{(2 + 3t, t, 6) | t \in \mathbb{R} \}$$
Is this right? I literally just copied the wording from my book.