Answer
The parallelogram with vertices $(0,0), (6,3), (12,1), (6,-2)$
Work Step by Step
Let us consider $u=\dfrac{x+3y}{5}; y=\dfrac{x-2y}{5}$
with $0 \leq u \leq 3$ and $0 \leq v \leq 2$
Re-arrange as:
$0 \leq \dfrac{x+3y}{5} \leq 3;\\0 \leq \dfrac{x-2y}{5} \leq 2$
This implies that
$0 \leq x+3y \leq 15; \\ 0 \leq x-2y \leq 10$
Hence, the parallelogram has vertices $(0,0), (6,3), (12,1), (6,-2)$
