On the coordinate plane, there is a parallelogram $\Gamma$, where two sides lie on lines parallel to $5x - y = 0$, and the other two sides lie on lines perpendicular to $3x - 2y = 0$. Let $Q$ be the intersection point of the two diagonals of $\Gamma$. It is known that $\Gamma$ has a vertex $P$ satisfying $\overrightarrow{PQ} = (10, -1)$. The area of $\Gamma$ is (11--1)(11--2)(11--3).