Points A($$\alpha$$, $$-$$3), B(2, 0) and C(1, $$\alpha$$) are collinear.

$$\therefore$$ Slope of AB = Slope of BC

$$ \Rightarrow {{0 + 3} \over {2 - \alpha }} = {{\alpha - 0} \over {1 - 2}}$$

$$ \Rightarrow - 3 = \alpha (2 - \alpha )$$

$$ \Rightarrow - 3 = 2\alpha - {\alpha ^2}$$

$$ \Rightarrow {\alpha ^2} - 2\alpha - 3 = 0$$

$$ \Rightarrow {\alpha ^2} - 3\alpha + \alpha - 3 = 0$$

$$ \Rightarrow \alpha (\alpha - 3) + 1(\alpha - 3) = 0$$

$$ \Rightarrow (\alpha + 1)(\alpha - 3) = 0$$

$$ \Rightarrow \alpha = - 1,\,3$$

Given, $${\alpha _1} < {\alpha _2}$$

$$\therefore$$ $${\alpha _1} = -1$$ and $${\alpha _2} = 3$$

$$\therefore$$ $$\left( {{\alpha _1},\,{\alpha _2}} \right) = ( - 1,\,3)$$

Now, equation of the line passing through ($$-$$1, 3) and making angle $${\pi \over 3}$$ with positive x-axis is

$$(y - {y_1}) = m(x - {x_1})$$

$$ \Rightarrow y - 3 = \left( {\tan {\pi \over 3}} \right)(x + 1)$$

$$ \Rightarrow y - 3 = \sqrt 3 (x + 1)$$

$$ \Rightarrow \sqrt 3 x - y + \sqrt 3 + 3 = 0$$