Charge on capacitor $C_1$

$=C_1\times 9=9 \mathrm{\mu C}$.

Now when the switch is thrown to the left side, charge distribution on left capacitors will be as given below,

For parallel combination along point $A$ and $B$, potential drop across it will be same,

$$\begin{gathered} \frac{Q}{C_1}=\frac{{\displaystyle q}}{{\displaystyle C_2}}+\frac{{\displaystyle q}}{{\displaystyle C_3}} \\ \Rightarrow Q=2q \end{gathered}$$

Also, total charge in the branches will be equal to the initial charge on the capacitor $C_1$. Therefore,

$$\begin{gathered} Q+q=9 \mathrm{\mu C} \\ \Rightarrow 2q+q=9 \mathrm{\mu C} \\ \Rightarrow q=3 \mathrm{\mu C} \end{gathered}$$