Answer
$10A+2B +2C$
Work Step by Step
The determinant for a $3 \times 3$ matrix can be found as:
$det = \begin{vmatrix}
a & b & c \\
d &e & f \\
g &h & i \\ \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)$
As per the problem,
$det=\begin{vmatrix}
A & B & C \\
1 & -2 & -3 \\
0 & 2 & -2 \\
\end{vmatrix}\\=A[(-2)(-2)-(-3)(2)]-B((1)(-2)- (-3) (0)]+C[(1)(2)- (-2) (0)] \\=10A+2B +2C$
