jmc

We begin by grouping terms in the denominator so that it resembles a two-term expression: $(1+\sqrt{2})-\sqrt{3}$. This suggests that our next step is to multiply both the numerator and the denominator of our original expression by $(1+\sqrt{2})+\sqrt{3}$ so that we have a difference of squares. Doing this, we have: $$\begin{array}{rl}\frac{1}{1+\sqrt{2}-\sqrt{3}}& =\frac{1}{(1+\sqrt{2})+\sqrt{3}}\times \frac{(1+\sqrt{2})+\sqrt{3}}{(1+\sqrt{2})-\sqrt{3}}\\ & =\frac{(1+\sqrt{2})+\sqrt{3}}{(1+\sqrt{2}{)}^2-(\sqrt{3}{)}^2}\\ & =\frac{1+\sqrt{2}+\sqrt{3}}{(1+2\sqrt{2}+2)-3}\\ & =\frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}.\end{array}$$We can then rationalize the denominator of this expression by multiplying both the numerator and denominator by $\sqrt{2}$ to get: $$\frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}=\frac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}+2+\sqrt{6}}{4}.$$Thus, $a=2$, $b=6$, and $c=4$, so we have $a+b+c=2+6+4=\boxed{12}$.