The given system of equations will have more than one solution if: \[ \begin{vmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & \lambda \end{vmatrix} = 0 \] Applying \( R_3 \to R_3 - (R_1 + 3R_2) \), we get: \[ \begin{vmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 0 & 0 & \lambda + 5 \end{vmatrix} = 0 \] Expanding along the third row: \[ 3[2(\lambda + 5) - 0] - (-1)(\lambda + 5 - 0) + 4(0 - 0) = 0 \] \[ 7\lambda + 35 = 0 \] \[ \lambda = \frac{-35}{7} \] \[ \lambda = -5 \]