jmc

The common denominator desired is $\sqrt{2}\cdot \sqrt{3}=\sqrt{6}$. So, this expression becomes $\frac{\sqrt{2}\cdot (\sqrt{2}\cdot \sqrt{3})+1\cdot \sqrt{3}+\sqrt{3}\cdot (\sqrt{2}\cdot \sqrt{3})+1\cdot \sqrt{2}}{\sqrt{6}}$. Simplifying this gives $\frac{2\sqrt{3}+\sqrt{3}+3\sqrt{2}+\sqrt{2}}{\sqrt{6}}=\frac{4\sqrt{2}+3\sqrt{3}}{\sqrt{6}}$. To rationalize, multiply numerator and denominator by $\sqrt{6}$ to get $\frac{4\sqrt{2}\sqrt{6}+3\sqrt{3}\sqrt{6}}{6}$. Simplifying yields $\frac{9\sqrt{2}+8\sqrt{3}}{6}$, so the desired sum is $9+8+6=\boxed{23}$.