Explanation

(a) Since,$C_{1}\ and\ C_{2}$ are parallel to their equivalent capacitance will be $(C_{1} + C_{2})$. Now, $\left( C_{1} + C_{2} \right)\ and\ C_{3}$ are in series, so the net equivalent capacitances of circuit.
$\frac{1}{C} = \frac{1}{C^{3}} + \frac{1}{C_{1} + C_{2}}$
$= \ \frac{C_{1} + C_{2} + C_{3}}{\left( C_{1} + C_{2} \right)C_{3}}$
$C = \ \frac{\left( C_{1} + C_{2} \right)C_{3}}{C_{1} + C_{2} + C_{3}}$
Since, $V$ is the voltage of the battery, so charge on this system
$q = CV$
$q = \ \frac{\left( C_{1} + C_{2} \right)C_{3}V}{C_{1} + C_{2} + C_{3}}$
If the capacitor $C_{3}$ breaks down then total equivalent capacitance
$C^{'} = \ C_{1} + C_{2}$
$\therefore$ New charge stored
$q^{'} = C'V$
$q' = \left( C_{1} + C_{2} \right)V$
Change in total charge
$\mathrm{\Delta}q = q^{'} - q$ $(\because q^{'} > q)$
$= \left( C_{1} + C_{2} \right)V - \frac{\left( C_{1} + C_{2} \right)C_{3}V}{C_{1} + C_{2} + C_{3}}\ \ $
$= \left( C_{1} + C_{2} \right)V\left\lbrack 1 - \frac{C_{3}}{C_{1} + C_{2} + C_{3}} \right\rbrack$
$\mathrm{\Delta}q = \ \left( C_{1} + C_{2} \right)V\left\lbrack 1 - \frac{C_{3}}{C_{1} + C_{2} + C_{3}}\ \right\rbrack$