Step 1
The matrix equation is $ \begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & -2 \\ \lambda & 1 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} \begin{bmatrix} 2 \\ 2 \\ 2 \end{bmatrix}$
$ AX = B$
The angmented matrix
$ [A,B] = \begin{bmatrix} 1 & 1 & 1 & 2 \\ 2 & 1 & -2 & 2 \\ \lambda & 1 & 4 & 2 \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & -1 & -4 & -2 \\ 0 & 1-\lambda & 4-\lambda & 2\lambda \end{bmatrix} R_2 \rightarrow R_2-2R_1$
$ R_3 \rightarrow R_3- \lambda R_1$
$ \sim \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & -1 & -4 & -2 \\ 0 & -\lambda & -\lambda & -2\lambda \end{bmatrix} R_3 \rightarrow R_3+R_2$
When $ \lambda = 0$ the last equivalent matrix becomes $ \begin{bmatrix} 1 & 1 & 1 & 2 \\ 0 & -1 & -4 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix}$ which is in echelon form.
There are two nonzero rows and $ \rho(A) = \rho(A,B)=2$
The system is consistent with infinitely many solutions.
When $ \lambda \neq 0, $ there are 3 nonzero rows and $ \rho(A)=\rho(A,B)=3$. The system has a unique solution.