Answer
This exchange of rows 1 and 2 reverses the sign of the determinant.
Work Step by Step
$a\begin{vmatrix}2&1\\5&6\end{vmatrix}-b\begin{vmatrix}3&1\\4&6\end{vmatrix}+c\begin{vmatrix}3&2\\4&5\end{vmatrix}=7a-14b+7c$
$3\begin{vmatrix}b&c\\5&6\end{vmatrix}-2\begin{vmatrix}a&c\\4&6\end{vmatrix}+1\begin{vmatrix}a&b\\4&5\end{vmatrix}=18b-15c-12a+8c+5a-4b=-7a+14b-7c=-(7a-14b+7c)$