Both $ a $ and $ b $ must be divisible by $ 6 $, so the choices for $ a $ are $$12, 42, 72, 102, 132, \ldots$$ and the choices for $ b $ are $$24, 54, 84, 114, 144, \ldots~.$$ We know that $ \mathop{\text{lcm}}[a,b]\cdot \gcd(a,b)=ab $ (since this identity holds for all positive integers $ a $ and $ b $). Therefore, $$\mathop{\text{lcm}}[a,b] = \frac{ab}{6},$$ so in order to minimize $ \mathop{\text{lcm}}[a,b] $, we should make $ ab $ as small as possible. But we can't take $ a=12 $ and $ b=24 $, because then $ \gcd(a,b) $ would be $ 12 $, not $ 6 $. The next best choice is either $ a=12,b=54 $ or $ a=42,b=24 $. Either of these pairs yields $ \gcd(a,b)=6 $ as desired, but the first choice, $ a=12 $ and $ b=54 $, yields a smaller product. Hence this is the optimal choice, and the smallest possible value for $ \mathop{\text{lcm}}[a,b] $ is $$\mathop{\text{lcm}}[12,54] = \frac{12\cdot 54}{6} = 2\cdot 54 = \boxed{108}.$$