Why parameterize systems of equations that contain "free variables"? As an example, let's say you have the system of equations given by

$$x+y+z = 1$$

$$2x+2y+2z = 2$$

$$3x+3y+3z=3$$

As the first few steps, we recognize that solutions to the first equation are solutions to all of them, so we parameterize the first, for instance by writing 

$$x=r$$

$$y=s$$

$$z = 1-r-s$$

Point being to show that we can insert arbitrary values into the parameters $r, s$, and produce a $z$ which gives us a guaranteed solution to the system above.

Question

Why go through the step of parameterizing the "free variables"? It always seemed like a redundant step to me. I've perused at least 7 Linear Algebra books hoping that one would give a reason, but they all just tell us to do it.
Parametrizing a space with coordinates from a know space gives you a more concrete description of the solution space. It can tell you the dimension of the solution space, it can give you explicit descriptions of what the solutions look like, and it can facilitate checking whether a given point is a solution or not.

In this particular case there is indeed no need to introduce new variables $r$ and $s$. You can simply parametrize the space of all solutions $V$ as $(x,y,1-x-y)$ with $x,y\in\Bbb{R}$. But introducing these new variables comes from the view that the parametrization is in fact a bijection
$$f:\ \Bbb{R}^2\ \longrightarrow\ V:\ (r,s)\ \longrightarrow\ (r,s,1-r-s).$$
From this point of view the parameters $r$ and $s$ are coordinates in a new space, the parametrizing space $\Bbb{R}^2$, and hence they get new names.

Now it turns out that for systems of linear equations over a field you can always choose a parametrization in terms of the coordinate functions (here in terms of $x$ and $y$). In this case these new parameter names seem redundant (and perhaps they are?). But outside the scope of linear algebra these are not at all redundant. For example, it is impossible to parametrize the rational points on the circle in terms of any of its coordinates; here it is necessary to introduce a new parameter.
The set $\{(x,y,z)\in \Bbb R^3 \; : x+y+z=1\}$ is a plane whose cartesian equation is $x+y+z=1$ and whose parametric equations are
$x=x, y=y, z=1-x-y$

or

$x=a, y=b, z=1-a-b$.

The parametric equations give the two vectors director of the plane.
in your case, the vectors are
$$u=(1,0,-1) \text{ and } v=(0,1,-1)$$