Answer
\begin{aligned} \operatorname{det}(A) &=\left|\begin{array}{ccc}{0} & {0} & {-3} \\ {0} & {4} & {3} \\ {-2} & {1} & {5}\end{array}\right| =-24 \end{aligned}
Work Step by Step
Given $$ A=\left[ \begin{array}{ccc}{0} & {0} & {-3} \\ {0} & {4} & {3} \\ {-2} & {1} & {5}\end{array} \right]$$
Since, if we have$$A=\left[\begin{array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{array}\right]$$
we get
\begin{aligned}\operatorname{det}(A)&=\left|\begin{array}{lll}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{array}\right|\\
&=a_{11}\left|\begin{array}{ll} {a_{22}} & {a_{23}} \\ {a_{32}} & {a_{33}}\end{array}\right|-a_{12}\left|\begin{array}{ll} {a_{21}} & {a_{23}} \\ {a_{31}} & {a_{33}}\end{array}\right|+a_{13}\left|\begin{array}{ll} {a_{21}} & {a_{22}} \\ {a_{31}} & {a_{32}}\end{array}\right|\\
&=a_{11}( a_{22} a_{33}-a_{23} a_{32})-a_{12}( a_{21} a_{33}-a_{23} a_{31})+a_{13}( a_{21} a_{32}-a_{22} a_{31})\\
&=a_{11} a_{22} a_{33}+a_{12} a_{23} a_{31}+a_{13} a_{21} a_{32}-a_{11} a_{23} a_{32}-a_{12} a_{21} a_{33}-a_{13} a_{22} a_{31}\\
\end{aligned}
Therefore, we get
\begin{aligned} \operatorname{det}(A) &=\left|\begin{array}{ccc}{0} & {0} & {-3} \\ {0} & {4} & {3} \\ {-2} & {1} & {5}\end{array}\right| \\ &=0 \cdot 4 \cdot 5+0 \cdot 3 \cdot(-2)+(-3) \cdot 0 \cdot 1-0 \cdot 3 \cdot(-1)-0 \cdot 0 \cdot 5-(-3) \cdot 4 \cdot(-2) \\ &=0-0-0+0-0-24\\
&=-24 \end{aligned}
