The equation for the line is

$$P = {{ - {P_0}} \over {{V_0}}}V + 3P$$

[slope $$=$$ $${{ - {P_0}} \over {{V_0}}},\,\,c = 3{P_0}$$ ]

$$P{V_0} + {P_0}V = 3{P_0}{V_0}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

But $$\,\,\,\,\,\,\,\,\,\,\,\,pv = nRT$$

$$\therefore$$ $$p = {{nRT} \over v}\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

From $$(i)$$ & $$(ii)$$ $${{nRT} \over v}{V_0} + {P_0}V = 3{P_0}{V_0}$$

$$\therefore$$ $$nRT{V_0} + {P_0}{V^2} = 3{P_0}{V_0}\,\,\,\,\,...\left( {iii} \right)$$

For temperature to be maximum $${{dT} \over {dv}} = 0$$

Differentiating e.q.$$(iii)$$ by $$'v'$$ we get

$$nR{V_0}{{dT} \over {dv}} + {P_0}\left( {2v} \right) = 3{P_0}{V_0}$$

$$\therefore$$ $$nR{V_0}{{dT} \over {dv}} = 3{P_0}{V_0} - 2{P_0}V$$

$${{dT} \over {dv}} = {{3{P_0}{V_0} - 2{P_0}V} \over {nR{V_0}}} = 0$$

$$V = {{3{V_0}} \over 2}$$

$$\therefore$$ $$p = {{3{P_0}} \over 2}\,\,\,\,\,\,\,\,\,$$ [From $$(i)$$ ]

$$\therefore$$ $${T_{\max }} = {{9{P_0}{V_0}} \over {4nR}}\,\,\,\,\,\,\,\,\,\,$$ [ From $$(ii)$$ ]