The charges on two metal spheres, before coming in contact, are given by

$$\begin{aligned} Q & =\sigma .4 \pi R^2 \\ Q_2 & =\sigma .4 \pi\left(2 R^2\right) \\ & =4\left(\sigma .4 \pi R^2\right)=4 Q_1 \end{aligned}$$

Let the charges on two metal spheres, after coming in contact becomes $Q_1^{\prime}$ and $Q_2^{\prime}$.

Now applying law of conservation of charges is given by

$$\begin{aligned} Q_1^{\prime}+Q_2^{\prime} & =Q_1+Q_2=5 Q_1 \\ & =5\left(\sigma \cdot 4 \pi R^2\right) \end{aligned}$$

After coming in contact, they acquire equal potentials. Therefore, we have

$$\frac{1}{4 \pi \varepsilon_0} \frac{Q_1^{\prime}}{R}=\frac{1}{4 \pi \varepsilon_0} \frac{Q_2^{\prime}}{R}$$

On solving, we get

$$\begin{array}{ll} \therefore & Q_1^{\prime}=\frac{5}{3}\left(\sigma .4 \pi R^2\right) \text { and } Q_2^{\prime}=\frac{10}{3}\left(\sigma .4 \pi R^2\right) \\ \therefore & \sigma_1=5 / 3 \sigma \text { and } \\ \therefore & \sigma_2=\frac{5}{6} \sigma \end{array}$$

In the circuit, when initially $K_1$ is closed and $K_2$ is open, the capacitors $C_1$ and $C_2$ acquires potential difference $V_1$ and $V_2$ respectively. So, we have

$$\begin{aligned} &\text { and }\\ &\begin{aligned} & V_1+V_2=E \\ & V_1+V_2=9 V \end{aligned} \end{aligned}$$

Also, in series combination ,

$$\begin{aligned} V & \propto 1 / C \\ V_1: V_2 & =1 / 6: 1 / 3 \end{aligned}$$

On solving

$$\begin{array}{ll} \Rightarrow & V_1=3 V \text { and } V_2=6 \mathrm{~V} \\ \therefore & Q_1=C_1 V_1=6 \mathrm{C} \times 3=18 \propto \mathrm{C} \\ & Q_2=9 \propto \mathrm{C} \text { and } Q_3=0 \end{array}$$

Then, $K_1$ was opened and $K_2$ was closed, the parallel combination of $C_2$ and $C_3$ is in series with $C_1$.

$$Q_2=Q_2^{\prime}+Q_3$$

and considering common potential of parallel combination as $V$, then we have

$$\begin{aligned} C_2 V+C_3 V & =Q_2 \\ \Rightarrow\quad V & =\frac{Q_2}{C_2+C_3}=(3 / 2) \mathrm{V} \\ \text{On solving,}\quad Q_2^{\prime} & =(9 / 2) \propto \mathrm{C} \\ \text{and}\quad Q_3 & =(9 / 2) \propto \mathrm{C} \end{aligned}$$

Let the point $P$ lies at a distance $x$ from the centre of the disk and take the plane of the disk to be perpendicular to the $\mathbf{x}$-axis. Let the disc is divided into a number of charged rings as shown in figure.

The electric potential of each ring, of radius $r$ and width $d r$, have charge $d q$ is given by

$$\sigma d A=\sigma 2 \pi r d r$$

and potential is given by

$$d V=\frac{k_e d q}{\sqrt{r^2+x^2}}=\frac{k_e \sigma 2 \pi r d r}{\sqrt{r^2+x^2}}$$

where $k_e=\frac{1}{4 \pi \varepsilon_0}$ the total electric potential at $P$, is given by

$$\begin{aligned} &\begin{aligned} & V=\pi k_e \sigma \int_0^a \frac{2 r d r}{\sqrt{r^2+x^2}}=\pi k_e \sigma \int_0^a\left(r^2+x^2\right)^{-1 / 2} 2 r d r \\ & V=2 \pi k_e \sigma\left[\left(x^2+a^2\right)^{1 / 2}-x\right] \end{aligned}\\ &\begin{aligned} \text{So, we have by substring}\quad k_e & =\frac{1}{4 \pi \varepsilon_0} \\ V & =\frac{1}{4 \pi \varepsilon_0} \frac{2 Q}{a^2}\left[\sqrt{x^2+a^2}-x\right] \end{aligned} \end{aligned}$$

Let any arbitrary point on the required plane is $(x, y, z)$. The two charges lies on $z$-axis at a separation of $2 d$.

The potential at the point $P$ due to two charges is given by

$$\begin{array}{ll} & \frac{q_1}{\sqrt{x^2+y^2+(z-d)^2}}+\frac{q_2}{\sqrt{x^2+y^2+(z+d)^2}}=0 \\ \therefore & \frac{q_1}{\sqrt{x^2+y^2+(z-d)^2}}=\frac{q_2}{\sqrt{x^2+y^2+(z+d)^2}} \end{array}$$

On squaring and simplifying, we get

$$x^2+y^2+z^2+\left[\frac{\left(q_1 / q_2\right)^2+1}{\left(q_1 / q_2\right)^2-1}\right](2 z d)+d^2-0$$

This is the equation of a sphere with centre at

$$\left(0,0,-2 d\left[\frac{q_1^2+q_2^2}{q_1^2-q_2^2}\right]\right)$$

Let third charge $+q$ is slightly displaced from mean position towards first charge. So, the total potential energy of the system is given by

$$\begin{aligned} U & =\frac{1}{4 \pi \varepsilon_0}\left\{\frac{-q^2}{(d-x)}+\frac{-q^2}{(d+x)}\right\} \\ U & =\frac{-q^2}{4 \pi \varepsilon_0} \frac{2 d}{\left(d^2-x^2\right)} \\ \frac{d U}{d x} & =\frac{-q^2 \cdot 2 d}{4 \pi \varepsilon_0} \cdot \frac{2 x}{\left(d^2-x^2\right)^2} \end{aligned}$$

The system will be in equilibrium, if

$$F=-\frac{d U}{d x}=0$$

On solving, $x=0$. So for, $+q$ charge to be in stable/unstable equilibrium, finding second derivative of PE.

$$\begin{aligned} \frac{d^2 U}{d x^2} & =\left(\frac{-2 d q^2}{4 \pi \varepsilon_0}\right)\left[\frac{2}{\left(d^2-x^2\right)^2}-\frac{8 x^2}{\left(d^2-x^2\right)^3}\right] \\ & =\left(\frac{-2 d q^2}{4 \pi \varepsilon_0}\right) \frac{1}{\left(d^2-x^2\right)^3}\left[2\left(d^2-x^2\right)^2-8 x^2\right] \\ \text{At}\quad x & =0 \\ \frac{d^2 U}{d x^2} & =\left(\frac{-2 d q^2}{4 \pi \varepsilon_0}\right)\left(\frac{1}{d^6}\right)\left(2 d^2\right), \text { which is }<0 \end{aligned}$$

This shows that system will be unstable equilibrium.