$$\sqrt{6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}}$$ =

(Hint:taking the cue from the options we should get the term as a square of 3 added terms.)

Now 

$$\sqrt{6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}}$$ = 

$$=\sqrt{\ \ \left(\left(1+2+3\right)+2.1.\sqrt{\ 2}+2.1.\sqrt{\ 3}+2.\sqrt{\ 2}.\sqrt{\ 3}\right)}$$

As $$\sqrt{\ \left(a+b+c\right)^2}=\sqrt{a^2+b^2+c^2+2.a.b+2.b.c+2.c.a\ \ }$$

On comparison, 

$$a=1\ ,\ b=\sqrt{\ 2\ },c\ =\sqrt{\ 3}$$

Hence,

$$\sqrt{6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}}$$ = $$1 +\sqrt{2} + \sqrt{3}$$ Answer