\(\mathbb{S}_1\):
\(\begin{eqnarray*} x + 2 y + z &=& 0\\ y - z &=& 2 \end{eqnarray*}\)
Insert a third equation to represent all possible values of \(z\).
\(\mathbb{S}_2\):
\(\begin{eqnarray*} x + 2 y + z &=& 0 &\quad E_1 \to -E_3 + E_1\\ y - z &=& 2 &\quad E_2 \to E_3 + E_2\\ z &=& t \end{eqnarray*}\)
\(\mathbb{S}_3\):
\(\begin{eqnarray*} x + 2 y ~\quad\quad &=& -t &\quad E_1 \to -2E_2 + E_1\\ y ~\quad\quad &=& 2 + t\\ z &=& t \\ \end{eqnarray*}\)
\(\mathbb{S}_4\):
\(\begin{eqnarray*} x~\quad\quad ~\quad\quad &=& -4 -3t\\ y ~\quad\quad &=& 2 + t\\ z &=& t \end{eqnarray*}\)
Thus the solution is every ordered number triple \(S=(-4-3t,2+t,t)=(-3,1,1)\cdot t+(-4,2,0)\) where \(t\) is a real number.
If this reminds you of the equation \(y=mx+b\) of a straight line, it is because the solution set is the set of all points \(S\) in an \(x,y,z\) coordinate system lying on the line whose vector equation is \(S=(-3,1,1)\cdot t+(-4,2,0)\). Recall that each of the equations in the system is the equation of a plane in the \(x,y,z\) coordinate system. This line is the line of intersection of the planes.
There are an infinite number points on that line, one for each value of \(t\), and the coordinates of each point represent one of the infinitely many solutions of this dependent system.