Question

For $C_{A}$ :

Let $\frac{\varepsilon_{0}\text{ }A}{\text{ }d} = C$

$$\therefore C_{A} = \frac{K_{1}C}{2} + \frac{K_{1}{\text{ }K}_{2}C}{K_{1} + K_{2}}$$

$$= K_{1}C\left\lbrack \frac{{\text{ }K}_{1} + 2{\text{ }K}_{2}}{2\left( {\text{ }K}_{1} + K_{2} \right)} \right\rbrack$$

For $C_{B}$ :

$$\begin{matrix} C_{B} & \ = \frac{K_{2}C}{2} + \frac{K_{1}{\text{ }K}_{2}C}{{\text{ }K}_{1} + K_{2}} \\ & \ = K_{2}C\left\lbrack \frac{{\text{ }K}_{1} + 2{\text{ }K}_{2}}{2\left( {\text{ }K}_{1} + K_{2} \right)} \right\rbrack \end{matrix}$$

For $C_{C}$ :

$$\begin{matrix} & C_{C} = \frac{2{\text{ }K}_{1}{\text{ }K}_{2}C}{\left( {\text{ }K}_{1} + K_{2} \right)} \\ & C_{A} > C_{C} > C_{B} \end{matrix}$$