jmc

Since 2, 3, and 7 are all primes, the denominator is in simplest radical form and we can't simplify it further. We attack this problem by getting rid of the square roots one step at a time. First we group the first two terms, and multiply numerator and denominator by the conjugate: $$\begin{array}{rl}\frac{1}{(\sqrt{2}+\sqrt{3})+\sqrt{7}}& =\frac{1}{(\sqrt{2}+\sqrt{3})+\sqrt{7}}\cdot \frac{(\sqrt{2}+\sqrt{3})-\sqrt{7}}{(\sqrt{2}+\sqrt{3})-\sqrt{7}}\\ & =\frac{(\sqrt{2}+\sqrt{3})-\sqrt{7}}{(\sqrt{2}+\sqrt{3}{)}^2-(\sqrt{7}{)}^2}\\ & =\frac{(\sqrt{2}+\sqrt{3})-\sqrt{7}}{2+2\sqrt{6}+3-7}\\ & =\frac{\sqrt{2}+\sqrt{3}-\sqrt{7}}{-2+2\sqrt{6}}\end{array}$$Now this is in a form we know how to deal with, and we can just multiply by the conjugate as usual: $$\begin{array}{rl}\frac{\sqrt{2}+\sqrt{3}-\sqrt{7}}{-2+2\sqrt{6}}& =\frac{\sqrt{2}+\sqrt{3}-\sqrt{7}}{-2+2\sqrt{6}}\cdot \frac{-2-2\sqrt{6}}{-2-2\sqrt{6}}\\ & =\frac{-2\sqrt{2}-2\sqrt{3}+2\sqrt{7}-2\sqrt{12}-2\sqrt{18}+2\sqrt{42}}{-20}\\ & =\frac{4\sqrt{2}+3\sqrt{3}-\sqrt{7}-\sqrt{42}}{10}.\end{array}$$This gives $A+B+C+D+E+F=4+3-1-1+42+10=\boxed{57}$.