\(\frac{1}{\sqrt{3}+\sqrt{2}+1}=\frac{\sqrt{3}+\sqrt{2}-1}{\left(\sqrt{3}+\sqrt{2}+1\right)\left(\sqrt{3}+\sqrt{2}-1\right)}\)=\(\frac{\sqrt{3}+\sqrt{2}-1}{\left(\sqrt{3}+\sqrt{2}\right)^2-1^2}=\frac{\sqrt{3}+\sqrt{2}-1}{5+2\sqrt{6}-1}=\frac{\sqrt{3}+\sqrt{2}-1}{4-2\sqrt{6}}\)