1. Given data

• Radius of sphere $S_1 = \dfrac{2R}{3}$
• Radius of sphere $S_2 = \dfrac{R}{3}$
• Initial charges: $q_1 = 12\,\mu C, \qquad q_2 = -3\,\mu C$

So, total charge is $Q_{\text{total}} = 12 + (-3) = 9\,\mu C$

2. Condition after connecting by a conducting wire

When two conducting spheres are connected by a wire and are far apart, charge redistributes until their potentials become equal.

For an isolated conducting sphere, $V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$

Thus after connection, $V_1 = V_2$

So, $\frac{q_1'}{r_1} = \frac{q_2'}{r_2}$

Substitute radii: $\frac{q_1'}{2R/3} = \frac{q_2'}{R/3}$

Cancel $R/3$: $\frac{q_1'}{2} = q_2'$

Hence, $q_1' : q_2' = 2 : 1$

3. Use conservation of total charge

Let final charges be $q_1' = 2x, \qquad q_2' = x$

Since total charge remains $9\,\mu C$, $2x + x = 9$ $3x = 9$ $x = 3$

Therefore, $q_1' = 6\,\mu C, \qquad q_2' = 3\,\mu C$

4. Match with options

This corresponds to:

• Option C: $6\,\mu C$ and $3\,\mu C$

5. Comparison with stored correct answer

Stored correct answer is C, which matches the derived result.