As $B \rightarrow A \rightarrow C$ is a closed cyclic process, we have

$\Delta U($ complete cycle $)=0$

So, by first law of thermodynamics,

we have $\Delta Q=\Delta W$

or $\Delta Q_{A B}+\Delta Q_{B C}+\Delta Q_{A C}=\Delta W$

$=\Delta W_{A B}+\Delta W_{B C}+\Delta W_{A C}$

$=$ Area enclosed under $p-V$ graph

$\Rightarrow \Delta Q_{A C}=-\left(\Delta Q_{A B}\right.\left.+\Delta Q_{B C}\right)+\frac{1}{2}$ $\left(V_2-V_1\right)\left(p_2-p_1\right)$ $=-\left(\frac{3}{2} n R \Delta T-p_1\left(V_2-V_1\right)\right.$ $\left.+\frac{3}{2} n R \Delta T-V_2\left(p_2-p_1\right)\right)$ $+\frac{1}{2}\left(V_2-V_1\right)\left(p_2-p_1\right)$

$\therefore \Delta Q_{A C}=2\left(p_2 V_2-\right.\left.p_1 V_1\right)$ $+\frac{1}{2}\left(V_2-V_1\right)\left(p_2-p_1\right)$