1. Given System of Equations: \[ \begin{cases} -x + y + z = 1 \\ -x - 2y + z = m \\ -x + y - 2z = n \end{cases} \] 2. Add the Three Equations: Adding the three equations, we get: \[ (-x + y + z) + (-x - 2y + z) + (-x + y - 2z) = 1 + m + n \] Simplifying: \[ -3x = 1 + m + n \] Given \(1 + m + n = 0\), we have: \[ -3x = 0 \Rightarrow x = 0 \] 3. Substitute \(x = 0\) into the Original Equations: Substituting \(x = 0\) into the original equations, we get: \[ \begin{cases} y + z = 1 \\ -2y + z = m \\ y - 2z = n \end{cases} \] 4. Solve for \(y\) and \(z\): Subtract the second equation from the first: \[ (y + z) - (-2y + z) = 1 - m \] \[ 3y = 1 - m \Rightarrow y = \frac{1 - m}{3} \] Subtract the third equation from the first: \[ (y + z) - (y - 2z) = 1 - n \] \[ 3z = 1 - n \Rightarrow z = \frac{1 - n}{3} \] 5. Determine the Number of Solutions: Since \(1 + m + n = 0\), \(m\) and \(n\) can vary as long as they satisfy this condition. This implies that \(y\) and \(z\) can take infinitely many values depending on the values of \(m\) and \(n\). Therefore, the given system of equations has infinitely many solutions: \[ \boxed{\text{infinitely many solutions}} \]