\(\mathbb{S}_1\):
\(\begin{eqnarray*} x + 2 y - ~~z &=& ~~~0 &\\ - 3 y + 3 z &=& ~~~0 &\quad E_2 \to -\left(\frac{1}{3}\right)\cdot E_2\\ ~~~7 y - ~~z &=& -6 & \end{eqnarray*}\)
\(\mathbb{S}_2\):
\(\begin{eqnarray*} x + 2 y - z &=& 0 &\\ y - z &=& 0 &\\ 7 y - z &=& -6 &\quad E_3 \to -7E_2 + E_3 \end{eqnarray*}\)
\(\mathbb{S}_3\):
\(\begin{eqnarray*} x + 2 y - z &=& 0\\ y - z &=& 0\\ 6 z &=& -6 \end{eqnarray*}\)
Solving the last equation, \(z = -1\). Substitution into the second equation, \(y = -1\). Substituting these two values into the first equation yields \(x = 1\).
Therefore, the solution is \((1,-1,-1)\).