Answer
$6$
Work Step by Step
We begin with an expansion down the fifth column, and continue with an expansion along the third row of the resultant matrix. For the $3\times3$ determinant, we expand down the first column. Only one term is non-vanishing in each of the first two expansions.
$
\begin{align}
\det A &=1\cdot\begin{vmatrix}6&3&2&4\\9&0&-4&1\\2&0&0&0\\4&2&3&2\end{vmatrix}=2\begin{vmatrix}3&2&4\\0&-4&1\\2&3&2\end{vmatrix}\\
&=2\cdot(3\begin{vmatrix}-4&1\\3&2\end{vmatrix}+2\begin{vmatrix}2&4\\-4&1\end{vmatrix})\\
&=6(-8-3)+4(2+16)=6
\end{align}
$
