Answer
$ 2A+12B -6C$
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,
$det=\begin{vmatrix}
A & B & C \\
0 & 2 & 4 \\
3 & 1 & 3 \\
\end{vmatrix}\\=A[(2)(3)-(4)(1)]-B((0)(3)- (4) (3)]+C[(0)(1)- (2) (3)] \\= 2A+12B -6C$
