Answer
$\det (A)=0$
Work Step by Step
We have the cofactor expansion theorem for the 1st row:
$\det (A)=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}$
with $C_{ij}=(-1)^{i+j}.M_{ij}$
So, we get:
$\det (A)=0C_{11}+(-2)C_{12}+1C_{13}$
$\det (A)=0\begin{vmatrix}
0 & -3 \\
3 & 0
\end{vmatrix}+(-2)\begin{vmatrix}
2 & -3 \\
-1 & 0
\end{vmatrix}+1\begin{vmatrix}
2 & 0 \\
-1 & 3
\end{vmatrix}$
Plug in the given values:
$\det (A)=0+2.(-3)+1.6$
$\det (A)=-6+6$
$\det (A)=0$
