If $$\left( 3 + \frac{1}{\sqrt{3}} + \frac{1}{3+\sqrt{3}} + \frac{1}{\sqrt{3} - 3} \right) \div (3+\sqrt{6}) = a + b\sqrt{6}$$ then what is the value of 3a-2b?

Correct option is A

Given:

​$$\left( 3 + \frac{1}{\sqrt{3}} + \frac{1}{3+\sqrt{3}} + \frac{1}{\sqrt{3} - 3} \right) \div (3+\sqrt{6}) = a + b\sqrt{6}$$ 

Solution:   

​$$\left( 3 + \frac{1}{\sqrt{3}} + \frac{1}{3+\sqrt{3}} + \frac{1}{\sqrt{3} - 3} \right) \div (3+\sqrt{6}) \\ \ \\ = \left( 3 + \frac{1}{\sqrt{3}} + \frac{1}{3+\sqrt{3}} + \frac{1}{-(-\sqrt{3} + 3)} \right) \div (3+\sqrt{6}) \\ \ \\ = \left( 3 + \frac{1}{\sqrt{3}} + \frac{1}{3+\sqrt{3}} - \frac{1}{(3-\sqrt{3} )} \right) \div (3+\sqrt{6})\\ \ \\ = \left( 3 + \frac{1}{\sqrt{3}} + \frac{(3-\sqrt{3})-(3+\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3} )} \right) \div (3+\sqrt{6})\\ \ \\ = \left( 3 + \frac{1}{\sqrt{3}} + \frac{(-2\sqrt{3})}{(9 -3)} \right) \div (3+\sqrt{6})\\ \ \\ = \left( 3 + \frac{\sqrt{3}}{3} - \frac{(\sqrt{3})}{(3)} \right) \div (3+\sqrt{6}) \\ \ \\ = \frac{3}{ 3+\sqrt{6}} \\ \ \\ = 3-\sqrt{6}$$​​

comparing this with $$a+b\sqrt6$$ 

we get, a = 3 and b = -1 

Now, value of 3a - 2b 

= 3(3) - 2(-1) = 9 + 2 = 11 

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