Answer
-18
Work Step by Step
Use the property that the determinant will not change when doing proper row or column operations.
$\begin{array}( \\|A|= \\ \\ \end{array}
\begin{vmatrix} 0 & 0 &4&6\\2 &1 &1&3\\2 &1 &2&3\\3 &0 &1&7 \end{vmatrix} \begin{array}( \\R_2-R_3\to R_2 \\ \\ \\\end{array}$
$\begin{array}( \\|A|= \\ \\ \end{array}
\begin{vmatrix} 0 & 0 &4&6\\0 &0 &-1&0\\2 &1 &2&3\\3 &0 &1&7 \end{vmatrix} \begin{array}( \ \\ =-(-1)\times\\ \\\end{array}\begin{vmatrix} 0 & 0 &6\\2 &1 &3\\3 &0 &7 \end{vmatrix} \begin{array}( \ \\ =(1)\times(6)\times\\ \\\end{array}\begin{vmatrix}2 &1 \\3 &0 \end{vmatrix}$
$|A|=6(-3)=-18$
