In the given circuit when, $\mathrm{K}_1$ is closed and $\mathrm{K}_2$ is open, the capacitors $\mathrm{C}_1$ and $\mathrm{C}_2$ will charge and potential difference develops accross $V_1$ and $V_2$ respectively. So, we have
$$
\mathrm{V}_1+\mathrm{V}_2=\mathrm{E}
$$
or $\quad V_1+V_2=9 \mathrm{~V}$
Also, in series combination,
$$
\begin{aligned}
&V=\frac{q}{C} \\
&\text { or, } \quad V \propto 1 / C \\
&\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\mathrm{C}_2}{\mathrm{C}_1} \\
&\mathrm{~V}_1: \mathrm{V}_2=\frac{1}{6}: \frac{1}{3} \\
&3 \mathrm{~V}_2=6 \mathrm{~V}_1,\left(\mathrm{~V}_2=2 \mathrm{~V}_1\right) \\
&
\end{aligned}
$$
Putting with the value of $\mathrm{V}_2$ in equation (i) then
$$
\begin{aligned}
\mathrm{V}_1+2 \mathrm{~V}_1 &=9 \\
3 \mathrm{~V}_1 &=9, \mathrm{~V}_1=3 \mathrm{~V} \\
\text { then, } \quad \quad \mathrm{V}_2 &=6 \mathrm{~V}
\end{aligned}
$$
so, $\quad \begin{array}{ll}\mathrm{Q}_1 & =\mathrm{C}_1 \mathrm{~V}_1=6 \mathrm{C} \times 3=18 \mu \mathrm{C} \\ \mathrm{Q}_2 & =\mathrm{C}_2 \mathrm{~V}_2=3 \times 6 \\ & =18 \mu \mathrm{C} \quad(\therefore \mathrm{a}=1 \mu \mathrm{f}) \\ \text { and } \quad \mathrm{Q}_3 & =0\end{array}$
Then, $\mathrm{K}_1$ was opened and $\mathrm{K}_2$ was closed, the parallel combination of $\mathrm{C}_2$ and $\mathrm{C}_3$ is in series with $\mathrm{C}_1$.
$$
\mathrm{Q}_2=\mathrm{Q}_2^{\prime}+\mathrm{Q}_3
$$
and considering common potential of parallel combination as $\mathrm{V}$. then we get :
$$
\begin{gathered}
\mathrm{C}_2 \mathrm{~V}+\mathrm{C}_3 \mathrm{~V}=\mathrm{Q}_2 \\
\mathrm{~V}=\frac{\mathrm{Q}_2}{\mathrm{C}_2+\mathrm{C}_3}=\frac{18}{3 \mathrm{C}+3 \mathrm{C}}=3 \mathrm{Volt}
\end{gathered}
$$
On solving, $\quad Q_2^{\prime}=C_2 V=3 C \times 3=9 \mu \mathrm{C}$ and
$$
\mathrm{Q}_3=\mathrm{C}_3 \mathrm{~V}=3 \mathrm{C} \times 3=9 \mu \mathrm{C}
$$