Answer
$\det (A)=0$
Work Step by Step
We have the cofactor expansion theorem is:
$\det (A)=a_{31}C_{31}+a_{32}C_{32}+a_{33}C_{33}$
with $C_{ij}=(-1)^{i+j}.M_{ij}$
to evaluate the given determinant of column 1:
$\det (A)=3C_{31}-(-5)C_{32}+0C_{33}$
$\det (A)=3\begin{vmatrix}
2 & -3 \\
0 & 5
\end{vmatrix}-(-5)\begin{vmatrix}
0 & -3 \\
-2 & 5
\end{vmatrix}+0\begin{vmatrix}
0 & 2 \\
-2 & 0
\end{vmatrix}$
Plug in the given values:
$\det (A)=3.10+5.(-6)+0$
$\det (A)=30-30+0$
$\det (A)=0$
