3a-6≥0 and 6-3a≥0
A≥2 and a≤2
So if a=2 is substituted into √(3a-6)+√(6-3a)+4=b, and b=4.
√(2a+3b)
=√(2×2+3×4)
=√(4+ 12)
=√ 16
=4