In describing the solution of a system of linear equations with many solutions, why do we use a free

Assume you and I (along with a few hundred other people) solve a system of linear equations independently and want to compare our results. If we have taken Gaussian elimination to reduced row-echelon form, and have expressed our answers with the free variables as parameters, we can immediately compare our answers; If there is any difference, then (at least) one of us is incorrect. If I used the free variables and you used the leading variables, we'll have to do some math before we can see if we got the same answer.
So, its

When there are more variables than equations, the solution of a system of linear equations is a line, plane, or other shape.
Suppose you have n variables and m equations, m Consider your example. You have 3 variables, which means that you have the whole 3D space of possible solutions:
${\displaystyle (x,y,z)}$
Each linear equation defines a plane in this space: ${\displaystyle ax+by+cz=d}$ is an equation for a plane. If you have two such equations, then the set of possible solutions is either a line, or the empty set (if two planes are parallel, e.g. ${\displaystyle x+y+z=1}$ and ${\displaystyle x+y+z=2}$). In your example, it is a line. You have already given two different ways to describe (parametrize) this line. Another way would be to fix x and solve for y and z:
${\displaystyle \{(x,\frac{7-x}{2},-\frac{x+1}{2})\mid x\in \mathbb{R}\}}$.
In fact, some variables do not need to be changed. Instead, you could use n-m equations. For example, let as add a new equation: ${\displaystyle y+z=2t}$. Therefore, you can solve the system of 3 equations with 3 variables $(x,y,z)$ as usually and obtain the following answer:
${\displaystyle \{(3-2t,t+2,t-2)\mid t\in \mathbb{R}\}}$
You can easily check that this set is exactly as any of the 3 previous ones: to get the first solution from this one just let ${\displaystyle t=z+2}$, the second one - ${\displaystyle t=y-2}$, and the third one ${\displaystyle t=\frac{3-x}{2}}$.