Answer
$6$
Work Step by Step
Although there are several promising strategies, we choose to begin by performing elementary row operations that create zeros in the fourth column of the matrix, and then expand down that column. We then "clean up" the resulting $3\times 3$ matrix, after which expansion down the first column leaves us with a single $2\times 2$ determinant to compute.
$\begin{vmatrix}2&5&4&1\\4&7&6&2\\6&-2&-4&0\\-6&7&7&0\end{vmatrix}=\begin{vmatrix}2&5&4&1\\0&-3&-2&0\\6&-2&-4&0\\-6&7&7&0\end{vmatrix}=-1\begin{vmatrix}0&-3&-2\\6&-2&-4\\-6&7&7\end{vmatrix}=-\begin{vmatrix}0&-3&-2\\6&-2&-4\\0&5&3\end{vmatrix}=- \left [-6\begin{vmatrix}-3&-2\\5&3\end{vmatrix} \right ]=6\cdot 1=6$
