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Let's look at this example more closely: $$\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$$ We find the solutions of this system by apply operations to the system to make a new system, aiming to end up with a very simple sort of system where we can see the solutions easily.
First replace equation (2) with $(2)-2\times (1)$. We'll call the resulting equations (1) and (2) again, although of course we end up with a different system of linear equations: $$\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$$ Now replace equation (1) with $(1)-3\times (2)$: $$\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$$ Notice that we can now easily rearrange (1) to find $x$ in terms of $z$, and we can rearrange (2) to find $y$ in terms of $z$. Since $z$ can take any value, we write $z=t$ where $t$ is a “free parameter” (which means $t$ can be any real number, or $t\in \mathbb{R}$). \begin{align*} x&=-16+5t\\ y&=7-2t\\ z&=t,\qquad t\in \mathbb{R}\end{align*} We can also write this in so-called “vector form”: \[ \begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix},\qquad t\in \mathbb{R}.\] This is the equation of the line where the two planes described by the original equations (1) and (2) intersect.
Note for each different value of $t$, we get a different solutions (that is, a different point on the line of intersection). For example, setting $t=0$ we see that $(-16,7,0)$ is a solution; setting $t=1.5$, we see that $(-16+1.5\times 5,7+1.5\times (-2),1.5) = (-8.5,4,1.5)$ is another solution, and so on. This works for any value $t\in\mathbb{R}$, and every solution may be written in this way.