Available resource ≥ remaining need of p1
Using 3 resources, p1 gets executed first. It frees 2 resources.
Available resources = 3 + 2 = 5
Available resource ≥ remaining need of p0
Using 5 resources, p0 gets executed first. It frees 5 resources.
Available resources = 5 + 5 = 10
Available resource ≥ remaining need of p2
Now, p2 can be executed easily. Hence, (p1, p0, p2) is a safe sequence.