Method 1
$ A = \begin{bmatrix} 0 & 1 & 2 & 1 \\ 2 & -3 & 0 & 1 \\ 1 & 1 & -1 & 0 \end{bmatrix}$ is a $3$ x $4$ matrix. $ \rho(A) \leq 3$
A has 4 3rd order minors.
Consider the minor $\begin{vmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & 1 & -1 \end{vmatrix} = 0-1(-2-0)+2(2+3)$
$ = 2+10=12 \neq 0$
A has atleast one 3rd order minor $ \neq 0$
$ \therefore \rho(A) = 3$
Method 2
$ A = \begin{bmatrix} 0 & 1 & 2 & 1 \\ 2 & -3 & 0 & 1 \\ 1 & 1 & -1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & 2 & 0 \\ 1 & -3 & 0 & 2 \\ 0 & 1 & -1 & 1 \end{bmatrix} (C_1 \leftrightarrow C_4)$
$ \sim \begin{bmatrix} 1 & 1 & 2 & 0 \\ 0 & -2 & -2 & 2 \\ 0 & 1 & -1 & 1 \end{bmatrix} R_2 \rightarrow R_2-R_1 \sim \begin{bmatrix} 1 & 1 & 2 & 0 \\ 0 & -2 & -2 & 2 \\ 0 & 0 & -4 & 4 \end{bmatrix} R_3 \rightarrow 2R_3+R_2$
The last equivalent matrix is in echelon form with 3 nonzero rows. $ \therefore \rho(A) = 3$