Step-by-step Solution:

To solve this system of equations, we are given the following: \[\] 1. \( x + 2y + 2z = 5 \) \[\] 2. \( x + 2y + 3z = 6 \) \[\] 3. \( x + 2y + \lambda z = \mu \) \[\] We need to find the conditions under which the system has infinitely many solutions. This can be determined by the rank of the augmented matrix and the coefficient matrix. \[\] The augmented matrix for this system is: \[ \left[ \begin{array}{ccc|c} 1 & 2 & 2 & 5 \\ 1 & 2 & 3 & 6 \\ 1 & 2 & \lambda & \mu \\ \end{array} \right] \] Now, let's perform the row operations: \[\] \( R_2 \rightarrow R_2 - R_1 \) \[\] \( R_3 \rightarrow R_3 - R_1 \) \[\] After performing these operations, the augmented matrix becomes: \[ \left[ \begin{array}{ccc|c} 1 & 2 & 2 & 5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & \lambda - 2 & \mu - 5 \\ \end{array} \right] \] For a system to have infinitely many solutions: \[ \text{rank(coefficient matrix)} = \text{rank(augmented matrix)} < \text{number of variables} \] one such case: \[ \lambda - 2 = 0 \quad \text{and} \quad \mu - 5 = 0 \] Thus, solving these conditions: \[ \lambda = 2 \quad \text{and} \quad \mu = 5 \] So, for many solutions (infinitely many), we need \( \lambda = 2 \) and \( \mu = 5 \).