∵a+b+c=3
∴2^(2a-3)·2^(3b-2)·2^(a+3c)
=2^(2a-3+3b-2+a+3c)
=2^(3a+3b+3c-5)
=(2³)^(a+b+c)÷2^5
=(2³)³÷2^5
=2^9÷2^5
=2^4
=16