Let $$A\left( {\alpha ,\,\frac{1}{\alpha }} \right),\,B\left( {\beta ,\,\frac{1}{\beta }} \right),\,C\left( {\gamma ,\,\frac{1}{\gamma }} \right)$$       be the vertices of a $$\Delta ABC,$$   where $$\alpha ,\,\beta $$  are the roots of the equation $${x^2} - 6{p_1}x + 2 = 0,\,\beta ,\,\gamma $$     are the roots of the equation $${x^2} - 6{p_2}x + 3 = 0$$    and $$\gamma ,\,\alpha $$  are the roots of the equation $${x^2} - 6{p_3}x + 6 = 0,\,{p_1},\,{p_2},\,{p_3}$$       being positive. Then, the coordinates of the centroid of $$\Delta ABC$$   is :

Solution :
It is given that $$\alpha ,\,\beta $$  are the roots of the equation $${x^2} - 6{p_1}x + 2 = 0$$
$$\therefore \,\alpha + \beta = 6{p_1},\,\alpha \beta = 2......\left( {\text{i}} \right)$$
$$\beta ,\,\gamma $$  are the roots of the equation $${x^2} - 6{p_2}x + 3 = 0$$
$$\therefore \,\beta + \gamma = 6{p_2},\,\beta \gamma = 3......\left( {{\text{ii}}} \right)$$
$$\gamma ,\,\alpha $$  are the roots of the equation $${x^2} - 6{p_3}x + 6 = 0$$
$$\therefore \,\gamma + \alpha = 6{p_3},\,\gamma \alpha = 6......\left( {{\text{iii}}} \right)$$
From Equations $$\left( {\text{i}} \right),\,\left( {{\text{ii}}} \right)$$   and $$\left( {{\text{iii}}} \right)$$  we get
$$\eqalign{ & \Rightarrow \alpha \beta \gamma = 6\,\,\,\,\,\left[ {\therefore \,\alpha ,\,\beta ,\,\gamma > 0} \right] \cr & {\text{Now, }}\alpha \beta = 2{\text{ and }}\alpha \beta \gamma = 6 \cr & \Rightarrow \gamma = 3 \cr & \beta \gamma = 3{\text{ }}\,\,{\text{and}}\,\,{\text{ }}\alpha \beta \gamma = 6 \cr & \alpha = 3,\,\,{\text{and }}\alpha = 6\alpha \beta \gamma = 6 \cr & \Rightarrow \beta = 1 \cr & \therefore \alpha + \beta = 6{p_1} \cr & \Rightarrow 3 = 6{p_1} \cr & \Rightarrow {p_1} = \frac{1}{2} \cr & \beta + \gamma = 6{p_2} \cr & \Rightarrow 4 = 6{p_2} \cr & \Rightarrow {p_2} = \frac{2}{3} \cr & {\text{and }}\,\gamma + \alpha = 6{p_3} \cr & \Rightarrow 5 = 6{p_3} \cr & \Rightarrow {p_3} = \frac{5}{6} \cr} $$
The coordinates of the centroid of triangle are
$$\eqalign{ & \left( {\frac{{\alpha + \beta + \gamma }}{3},\,\frac{1}{3}\left( {\frac{1}{\alpha } + \frac{1}{\beta } + \frac{1}{\gamma }} \right)} \right) \cr & {\text{or }}\left( {\frac{6}{3},\,\frac{1}{3}\left( {\frac{1}{2} + 1 + \frac{1}{3}} \right)} \right) \cr & {\text{or }}\left( {2,\,\frac{{11}}{{18}}} \right) \cr} $$