To solve this problem, we need to determine the steady-state error, \(e(t)\), for a unit step input in a feedback control system.

1. Given that the transfer function of the system is \(G(s) = \frac{6}{s(s+1)(s+2)}\).
2. The input \(r(t)\) is a unit step function, so \(R(s) = \frac{1}{s}\).
3. The error signal is given by \(E(s) = R(s) - Y(s)\). In steady-state conditions, \(\lim_{s \to 0} sE(s)\) can be analyzed using the Final Value Theorem.
4. The closed-loop transfer function is: \(T(s) = \frac{G(s)}{1 + G(s)}\).
5. Substituting the given \(G(s)\): \(T(s) = \frac{\frac{6}{s(s+1)(s+2)}}{1 + \frac{6}{s(s+1)(s+2)}}\).
6. Simplifying this: \(T(s) = \frac{6}{s^3 + 3s^2 + 2s + 6}\).
7. For a unit step input: \(Y(s) = T(s) \times R(s) = \frac{6}{s(s^3 + 3s^2 + 2s + 6)}\).
8. The error function is: \(E(s) = \frac{1}{s} - \frac{6}{s(s^3 + 3s^2 + 2s + 6)}\).
9. Applying the Final Value Theorem to find steady-state error: \(\lim_{t \to \infty} e(t) = \lim_{s \to 0} sE(s) = \lim_{s \to 0} \left(1 - \frac{6}{s^2 + 3s + 2}\right)\).
10. Analyzing reveals no finite limit due to pole at the origin; thus, it implies oscillations or instability.

Hence, the correct answer is that the steady-state error \(e(t)\) does not exist, and \(e(t)\) is oscillatory.

The correct answer is: \(\lim_{t\to\infty} e(t)\) does not exist, \(e(t)\) is oscillatory.