Answer
$6$
Work Step by Step
To compute the determinant efficiently, we expand the original matrix down the second column, the resulting matrix across the second row, and the third ($3\times3$) matrix down the first column. Only a single term is non-vanishing in each of the first two expansions. Note the initial negative sign because the sum of the indices of the $(3,2)$ position is odd.
$
\begin{align}
\det A &=-3\begin{vmatrix}4&-7&3&-5\\0&2&0&0\\5&5&2&-3\\0&9&-1&2\end{vmatrix}
=-3\cdot2\begin{vmatrix}4&3&-5\\5&2&-3\\0&-1&2\end{vmatrix}\\
&=-6\cdot(4\begin{vmatrix}2&-3\\-1&2\end{vmatrix}-5\begin{vmatrix}3&-5\\-1&2\end{vmatrix})\\
&=-6\cdot(4-5)=6
\end{align}
$