The simplified value of the following expression is: $$\frac{1}{\sqrt{11-2\sqrt{30}}}-\frac{3}{\sqrt{7-2\sqrt{10}}}-\frac{4}{\sqrt{8+4\sqrt{3}}}$$

Using, $$a^2+b^2+ab=(a+b)^2$$

=> $$\sqrt{11-2\sqrt{30}}=\sqrt{(\sqrt6)^2+(\sqrt5)^2-2\sqrt6\sqrt5}=(\sqrt6-\sqrt5)$$

Similarly, $$\sqrt{7-2\sqrt{10}}=(\sqrt5-\sqrt2)$$

and $$\sqrt{8+4\sqrt3}=\sqrt{8+2\sqrt{12}}=(\sqrt6+\sqrt2)$$

To find : $$\frac{1}{\sqrt{11-2\sqrt{30}}}-\frac{3}{\sqrt{7-2\sqrt{10}}}-\frac{4}{\sqrt{8+4\sqrt{3}}}$$

= $$\frac{1}{(\sqrt6-\sqrt5)}-\frac{3}{(\sqrt5-\sqrt2)}-\frac{4}{(\sqrt6+\sqrt2)}$$

Rationalizing the denominator, we get :

= $$[\frac{1}{\sqrt6-\sqrt5}\times\frac{\sqrt6+\sqrt5}{\sqrt6+\sqrt5}]-[\frac{3}{\sqrt5-\sqrt2}\times\frac{\sqrt5+\sqrt2}{\sqrt5+\sqrt2}]-[\frac{4}{\sqrt6+\sqrt2}\times\frac{\sqrt6-\sqrt2}{\sqrt6-\sqrt2}]$$

= $$(\sqrt6+\sqrt5)-(\sqrt5+\sqrt2)-(\sqrt6-\sqrt2)$$

= $$0$$

=> Ans - (A)