$\frac{1}{C_{e q}}=\frac{1}{C_{3}}+\frac{1}{C_{1}+C_{2}}$

(since $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are in parallel and which is in series with ${\mathrm{C}}_{3}$ ).

ie, $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}{\mathrm{C}_{3}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}$

$\therefore C_{e q}=\frac{C_{3}\left(C_{1}+C_{2}\right)}{C_{1}+C_{2}+C_{3}}$

since $V$ is the voltage of battery, charge, $\mathrm{q}=\mathrm{C}_{\mathrm{eq}} \mathrm{V}$

$=\frac{\mathrm{C}_{3}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}}{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}$

If the capacitor $\mathrm{C}_{3}$ breaks down, then effective capacitance, $C_{e q}^{\prime}=C_{1}+C_{2}$

New charge $q'=$ $\mathrm{C}_{\mathrm{eq}}^{\prime} \mathrm{V}=\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}$

Change in total charge $=q^{\prime}-q$

$=\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}-\frac{\mathrm{C}_{3}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}}{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}$

$=\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \mathrm{V}\left[1-\frac{\mathrm{C}_{3}}{\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}}\right]$