Answer
$20$
Work Step by Step
With a single row exchange, we are fortunate to find that expanding down fourth column of the $4\times 4$ matrix and across the second row of the $3\times 3$ matrix matrix reduces the original determinant of order 4 to a determinant of order 2.
$\begin{vmatrix}1&5&4&1\\0&-2&-4&0\\3&5&4&1\\-6&5&5&0\end{vmatrix}=\begin{vmatrix}1&5&4&1\\0&-2&-4&0\\2&0&0&0\\-6&5&5&0\end{vmatrix}=-1\begin{vmatrix}0&-2&-4\\2&0&0\\-6&5&5\end{vmatrix}=- \left [-2\begin{vmatrix}-2&-4\\5&5\end{vmatrix} \right ]=2\cdot 10=20$