Convert the given matrix in Row-Echelon form.
$\begin{bmatrix} 0 & 1 & t \\ 2 & t & -1 \\ 2 & 2 & 0 \end{bmatrix} \implies \begin{bmatrix} 2 & 2 & 0 \\ 2 & t & -1 \\ 0 & 1 & t \end{bmatrix} \implies \begin{bmatrix} 2 & 2 & 0 \\ 0 & 2 - t & 1 \\ 0 & 1 & t \end{bmatrix}$
Now, to make the second and third rows equal, we should have $$2 - t = t \implies t = 1$$
If $t = 1$, then rank of matrix will be $2$ since there are only two linearly independent rows in the matrix.
For all other values of $t$, the rank will be $3$.
So, Option $(C)$ is correct.