If $$\frac{4}{1 + \sqrt2 + \sqrt 3} = a + b\sqrt 2 + c \sqrt 3 - d\sqrt 6,$$ where a, b, c, d are natural numbers, then the value of a + b + c + d is:

Given,  $$\frac{4}{1 + \sqrt2 + \sqrt 3} = a + b\sqrt 2 + c \sqrt 3 - d\sqrt 6$$

$$=$$>  $$\frac{4}{1+\sqrt{2}+\sqrt{3}}\times\frac{1+\sqrt{2}-\sqrt{3}}{1+\sqrt{2}-\sqrt{3}}\ =a+b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$$

$$=$$>  $$\frac{4\left(1+\sqrt{2}-\sqrt{3}\right)}{\left(1+\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2}=a+b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$$

$$=$$>  $$\frac{4\left(1+\sqrt{2}-\sqrt{3}\right)}{1+2+2\sqrt{2}-3}=a+b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$$

$$=$$>  $$\frac{4\left(1+\sqrt{2}-\sqrt{3}\right)}{2\sqrt{2}}=a+b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$$

$$=$$>  $$\frac{4\left(1+\sqrt{2}-\sqrt{3}\right)}{2\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=a+b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$$

$$=$$>  $$\frac{4\sqrt{2}\left(1+\sqrt{2}-\sqrt{3}\right)}{2\left(\sqrt{2}\right)^2}=a+b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$$

$$=$$>  $$\frac{4\sqrt{2}\left(1+\sqrt{2}-\sqrt{3}\right)}{2\times\ 2}=a+b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$$

$$=$$>  $$\sqrt{2}\left(1+\sqrt{2}-\sqrt{3}\right)=a+b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$$

$$=$$>  $$\sqrt{2}+2-\sqrt{6}=a+b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$$

$$=$$>  $$2+\sqrt{2}-\sqrt{6}=a+b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$$

Comparing both sides

a=2, b=1, c=0, d=1

$$\therefore\ $$a + b + c + d = 2 + 1 + 0 + 1 = 4

Hence, the correct answer is Option D