Solution

Given that,
$$\alpha, \beta, \gamma$$ are the roots of $$x^3 - 11x^2 + 36x - 36 = 0$$
Here, $$a = 1, b= -11, c = 36, d = -36$$

$$\alpha + \beta + \gamma = \dfrac{-b}{a} = 11$$

$$\alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{c}{a} = 36$$

$$\alpha \beta \gamma = \dfrac{-d}{a} = 36$$

$$\because \dfrac{2}{\beta} = \dfrac{1}{\alpha} + \dfrac{1}{\gamma}$$

$$\implies \dfrac{2}{\beta} = \dfrac{\alpha + \gamma}{\alpha \gamma}$$

$$\implies 2\alpha \gamma= \alpha \beta + \beta \gamma$$

$$\implies 3\alpha \gamma= \alpha \beta + \beta \gamma + \alpha \gamma = 36$$

$$\implies \alpha \gamma = \dfrac{36}{3} = 12$$....................(i)

Now, $$\beta = \dfrac{\alpha \beta \gamma}{\alpha \gamma} = \dfrac{36}{12} = 3$$

$$\implies \alpha + \gamma = 11 - 3 = 8$$................................(ii)

From (i) & (ii) ,
$$\alpha, \beta$$ are $$6$$ and $$2$$

Hence, $$\alpha = 6$$
$$\beta = 3$$
$$\gamma = 2$$