Given that 2/a + 1/b = 9 and 4/a^2 + 1/b^2 = 57, find the value of 3ab.
\(\left(\frac{2}{a}+\frac{1}{b}\right)^2=\frac{4}{a^2}+\frac{1}{b^2}+\frac{4}{ab}=81\rightarrow 57+\frac{4}{ab}=81\rightarrow \frac{4}{ab}=24\rightarrow \frac{1}{6}=ab\rightarrow\boxed{3ab=\frac{1}{2}}\)
+605 
