Step 1: Initial charge on \( C_1 \) at \( t = 0 \) is:
\[q_1 = C_1 \cdot V_1 = 6 \, \mu F \cdot 5 \, \text{V} = 30 \, \mu C\]
Step 2: Upon closing switch 'S', charge redistributes between capacitors \( C_1 \) and \( C_2 \) until equilibrium. At equilibrium, the potential difference across both capacitors is equal, denoted as \( V_c \).
\[q_1 = C_1 \cdot V_c \quad \text{and} \quad q_2 = C_2 \cdot V_c\]
Charge conservation dictates:
\[q_1 + q_2 = 30 \, \mu C\]
Substituting the expressions for \( q_1 \) and \( q_2 \):
\[C_1 \cdot V_c + C_2 \cdot V_c = 30 \, \mu C\]
\[V_c \cdot (C_1 + C_2) = 30 \, \mu C\]
Solving for \( V_c \):
\[V_c = \frac{30 \, \mu C}{C_1 + C_2} = \frac{30 \, \mu C}{6 \, \mu F + 12 \, \mu F} = \frac{30 \, \mu C}{18 \, \mu F} = 1.67 \, \text{V}\]
Step 3: Final charges on the capacitors are:
\[q_1 = C_1 \cdot V_c = 6 \, \mu F \cdot 1.67 \, \text{V} = 10 \, \mu C\]
\[q_2 = C_2 \cdot V_c = 12 \, \mu F \cdot 1.67 \, \text{V} = 20 \, \mu C\]
Therefore, \( q_1 = 10 \, \mu C \) and \( q_2 = 20 \, \mu C \), corresponding to option (3).