$\begin{bmatrix} 1 &4 &8 &7\\ 0 &0& 3 &0\\ 4 &2& 3 &1\\ 3 &12 &24 &21 \end{bmatrix} = 3\begin{bmatrix} 1 &4 &8 &7\\ 0 &0& 3 &0\\ 4 &2& 3 &1\\ 1 &4 &8 &7 \end{bmatrix} $
$R_1$ and $R_4$ are the same and hence we can remove $R_4$ making the rank surely less than $4$.
$\text{Taking 3 out from $R_2$} \implies 9\begin{bmatrix} 1 &4 &8 &7\\ 0 &0& 1 &0\\ 4 &2& 3 &1 \end{bmatrix} $
${R_{1} \leftarrow R_{1}-8R_{2} \atop{R_{3} \leftarrow R_{3}-3R_{2}} } \implies 9\begin{bmatrix} 1 &4 &0 &7\\ 0 &0& 1 &0\\ 4 &2& 0 &1 \end{bmatrix} $
None of the rows are linearly independent (we cannot make any of them all $0's$.
So, Rank will be $\textbf{3}$.
(A) is correct option!