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Its evident that we have to apply $\color{#20A900}{(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca}$ to all the three numerators.
One can easily observe that terms in the numerator of the form $a^2$ add up while of the form $2ab$ add up to zero. Also we see squares of $2\sqrt2,2\sqrt3,2\sqrt5$ appear twice while squares of $\sqrt2,\sqrt5,\sqrt3$ appear only once. Hence answer is: $\dfrac{2((2\sqrt2)^2+(2\sqrt3)^2+(2\sqrt5)^2)+(\sqrt2)^2+(\sqrt3)^2+(\sqrt5)^2}{9} \\\Large=\boxed{\color{#007fff}{10}}$

Let terms inside brackets be a, b, c respectively. $a+b+c=\sqrt{2}+\sqrt{3}+\sqrt{5}$ So $a^2+b^2+c^2=2+3+5=10$