When you have more variables than equations, the solution of the system of linear equations is not just a point, but a line, or a plane etc.
Suppose you have $n$ variables and $m$ equations, $m<n$. You may "fix" extra $n-m$ variables and consider those just as numbers. By doing this, you essentially reduce the number of variables to $m$, equal to the number of equations, and express all other variables in terms of constants and fixed variables (parameters). And it does not matter which variables you fix and use as parameters (as long as each equation still contains a non-fixed variable): you are going to get the same set of solutions, just parametrized in different ways.
Consider your example. You have 3 variables, which means that you have the whole 3D space of possible solutions: $$(x,y,z)$$. Each linear equation defines a plane in this space: $ax+by+cz=d$ 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. $x+y+z=1$ and $x+y+z=2$). 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 $z$: $\{(x,\frac{7-x}{2},-\frac{x+1}{2})|x\in\mathbb{R}\}$.
In fact, you do not have to fix some variables. Instead, you may add $n-m$ equations (of course, parametrized by new external variables). For example, let as add a new equation: $y+z=2t$. Now we can solve the system of 3 equations with 3 variables ($x,y,z$) as usually and obtain the following answer: $\{(3-2t,t+2,t-2)|t\in\mathbb{R}\}$. 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 $t=z+2$, the second one -- $t=y-2$, and the third one $t=\frac{3-x}{2}$.
