$$\frac{1}{1 - \sqrt{2} + \sqrt{3}} + \frac{1}{1 - \sqrt{2} - \sqrt{3}} - \frac{2}{1 + \sqrt{2} - \sqrt{3}} + \frac{3}{\sqrt{2}}$$ equals

Expression : $$\frac{1}{1 - \sqrt{2} + \sqrt{3}} + \frac{1}{1 - \sqrt{2} - \sqrt{3}} - \frac{2}{1 + \sqrt{2} - \sqrt{3}} + \frac{3}{\sqrt{2}}$$

= $$[\frac{1}{1-\sqrt2+\sqrt3}\times\frac{(1-\sqrt2)-\sqrt3}{(1-\sqrt2)-\sqrt3}]+[\frac{1}{1-\sqrt2-\sqrt3}\times\frac{(1-\sqrt2)+\sqrt3}{(1-\sqrt2)+\sqrt3}]-2[\frac{1}{1+\sqrt2-\sqrt3}\times\frac{(1+\sqrt2)+\sqrt3}{(1+\sqrt2)+\sqrt3}]+[\frac{3}{\sqrt2}]$$

= $$[\frac{1-\sqrt2-\sqrt3}{(1-\sqrt2)^2-(\sqrt3)^2}]+[\frac{1-\sqrt2+\sqrt3}{(1-\sqrt2)^2-(\sqrt3)^2}]-2[\frac{1+\sqrt2+\sqrt3}{(1+\sqrt2)^2-(\sqrt3)^2}]+[\frac{3}{\sqrt2}]$$

= $$[\frac{1-\sqrt2-\sqrt3}{1-2\sqrt2+2-3}]+[\frac{1-\sqrt2+\sqrt3}{1-2\sqrt2+2-3}]-2[\frac{1+\sqrt2+\sqrt3}{1+2\sqrt2+2-3}]+[\frac{3}{\sqrt2}]$$

= $$[\frac{1-\sqrt2-\sqrt3}{-2\sqrt2}]+[\frac{1-\sqrt2+\sqrt3}{-2\sqrt2}]-2[\frac{1+\sqrt2+\sqrt3}{2\sqrt2}]+[\frac{6}{2\sqrt2}]$$

= $$[\frac{-1+\sqrt2+\sqrt3}{2\sqrt2}]+[\frac{-1+\sqrt2-\sqrt3}{2\sqrt2}]+[\frac{-2-2\sqrt2-2\sqrt3}{2\sqrt2}]+[\frac{6}{2\sqrt2}]$$

= $$\frac{1}{2\sqrt2}[(-1-1-2+6)+(\sqrt2+\sqrt2-2\sqrt2)+(\sqrt3-\sqrt3-2\sqrt3)]$$

= $$\frac{1}{2\sqrt2}(2-2\sqrt3)$$

= $$\frac{1-\sqrt3}{\sqrt2}$$

= $$\frac{1}{2}(\sqrt2-\sqrt6)$$

=> Ans - (D)