Answer
$-10$
Work Step by Step
We reduce the matrix to an echelon form by a series of row replacement operations and a single row swap.
$\begin{vmatrix}1&3&2&-4\\0&1&2&-5\\2&7&6&-3\\-3&-10&-7&2\end{vmatrix}=\begin{vmatrix}1&3&2&-4\\0&1&2&-5\\0&1&2&5\\0&-1&-1&-10\end{vmatrix}=\begin{vmatrix}1&3&2&-4\\0&1&2&-5\\0&0&0&10\\0&0&1&-15\end{vmatrix}=-\begin{vmatrix}1&3&2&-4\\0&1&2&-5\\0&0&1&-15\\0&0&0&-4\end{vmatrix}=-(1\cdot 1\cdot 1\cdot 10)=-10$
Recall that, when swapping two rows in a matrix, the determinant reverses sign; hence, we multiply by $-1$ in the ante-penultimate step.
