Two isolated conducting spheres $$S_1$$ and $$S_2$$ of radius $$\frac{2}{3}R$$ and $$\frac{1}{3}R$$ have 12 $$\mu C$$ and $$-3$$ $$\mu C$$ charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on $$S_1$$ and $$S_2$$ are respectively:

Initially sphere $$S_1$$ has radius $$r_1 = \dfrac{2}{3}R$$ and carries charge $$Q_1 = 12\,\mu\text{C}$$, whereas sphere $$S_2$$ has radius $$r_2 = \dfrac{1}{3}R$$ and carries charge $$Q_2 = -3\,\mu\text{C}$$. They are far apart, so each sphere’s potential with respect to infinity is produced only by its own charge, namely $$V = k\dfrac{Q}{r}$$ where $$k = \dfrac{1}{4\pi\varepsilon_0}$$.

When the two spheres are joined by a long, thin conducting wire, charge can flow freely until both spheres reach the same electric potential. Let the final charges be $$q_1$$ on $$S_1$$ and $$q_2$$ on $$S_2$$. Two fundamental conditions now govern the situation:

1. Conservation of charge (no charge is lost to the surroundings):

$$q_1 + q_2 = Q_1 + Q_2 = 12\,\mu\text{C} + (-3\,\mu\text{C}) = 9\,\mu\text{C}.$$

2. Equality of potentials, because the conducting wire forces the potentials to be identical:

$$V_1 = V_2 \quad\Longrightarrow\quad k\dfrac{q_1}{r_1} = k\dfrac{q_2}{r_2}.$$

The constant $$k$$ cancels on both sides, leaving

$$\dfrac{q_1}{r_1} = \dfrac{q_2}{r_2}.$$

Now substitute the given radii:

$$\dfrac{q_1}{\dfrac{2}{3}R} = \dfrac{q_2}{\dfrac{1}{3}R}.$$

To remove the complex denominators multiply numerator and denominator appropriately:

$$\dfrac{3q_1}{2R} = \dfrac{3q_2}{R}.$$

Cancel the common factor $$\dfrac{3}{R}$$ from both sides:

$$\frac{q_1}{2} = q_2.$$

Hence the simple relation between the final charges is

$$q_1 = 2q_2.$$

Insert this relation into the conservation equation $$q_1 + q_2 = 9\,\mu\text{C}$$:

$$2q_2 + q_2 = 9\,\mu\text{C}$$

$$3q_2 = 9\,\mu\text{C}$$

$$q_2 = \dfrac{9\,\mu\text{C}}{3} = 3\,\mu\text{C}.$$

Now find $$q_1$$ using $$q_1 = 2q_2$$:

$$q_1 = 2 \times 3\,\mu\text{C} = 6\,\mu\text{C}.$$

Therefore, after equilibrium is reached, sphere $$S_1$$ carries $$6\,\mu\text{C}$$ and sphere $$S_2$$ carries $$3\,\mu\text{C}$$.

Hence, the correct answer is Option D.