$$\because$$ $$f(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {f\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda ,\,x > 0} $$

On differentiating both sides w.r.t., x, we get

$$f'(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {{{{\lambda ^2}} \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda } $$

$$f'(x) = {1 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {\lambda \,.\,{{2\lambda } \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda } $$

$$\therefore$$ $$\sqrt 3 f'(x) = \left[ {{\lambda \over x}\,.\,f\left( {{{{\lambda ^2}x} \over 3}} \right)} \right]_0^{\sqrt 3 } - \int\limits_0^{\sqrt 3 } {{1 \over x}f\left( {{{{\lambda ^2}x} \over 3}} \right)dx} $$

$$\sqrt 3 x\,f'(x) = \sqrt 3 f(x) - {{\sqrt 3 } \over 2}f(x)$$

$$xf'(x) = {{f(x)} \over 2}$$

On integrating we get : $$\ln y = {1 \over 2}\ln x + \ln c$$

$$\because$$ $$f(1) = \sqrt 3 $$ then $$c = \sqrt 3 $$

$$\therefore$$ ($$\alpha$$, 6) lies on

$$\therefore$$ $$y = \sqrt {3x} $$

$$\therefore$$ $$6 = \sqrt {3\alpha } \Rightarrow \alpha = 12$$.