Answer
$4$
Work Step by Step
We can expand along the first column, then the determinant equals the determinant of the $3\times3$ matrix in the lower right "corner".
We know that for a matrix
\[
\left[\begin{array}{rrr}
a & b & c \\
d &e & f \\
g &h & i \\
\end{array} \right]
\]
the determinant, $D=a(ei-fh)-b(di-fg)+c(dh-eg).$
Hence here $D=3(4\cdot1-0\cdot0)-2((-2)\cdot1-0\cdot3)+1(-2\cdot0-4\cdot3)=3
\cdot4-2\cdot(-2)+1\cdot-12=12+4-12=4.$
