On a plane, Bob chooses 3 points $A_0$, $B_0$, $C_0$ (not necessarily distinct) such that $A_0B_0+B_0C_0+C_0A_0=1$. Then he chooses points $A_1$, $B_1$, $C_1$ (not necessarily distinct) in such a way that $A_1B_1=A_0B_0$ and $B_1C_1=B_0C_0$.
Next he chooses points $A_2$, $B_2$, $C_2$ as a permutation of points $A_1$, $B_1$, $C_1$. Finally, Bob chooses points $A_3$, $B_3$, $C_3$ (not necessarily distinct) in such a way that $A_3B_3=A_2B_2$ and $B_3C_3=B_2C_2$. What are the smallest and the greatest possible values of $A_3B_3+B_3C_3+C_3A_3$ Bob can obtain?
1. **Initial Setup and Given Conditions:**
   - Bob chooses points \( A_0, B_0, C_0 \) such that \( A_0B_0 + B_0C_0 + C_0A_0 = 1 \).
   - Bob then chooses points \( A_1, B_1, C_1 \) such that \( A_1B_1 = A_0B_0 \) and \( B_1C_1 = B_0C_0 \).
   - Next, Bob permutes \( A_1, B_1, C_1 \) to get \( A_2, B_2, C_2 \).
   - Finally, Bob chooses points \( A_3, B_3, C_3 \) such that \( A_3B_3 = A_2B_2 \) and \( B_3C_3 = B_2C_2 \).

2. **Objective:**
   - Determine the smallest and greatest possible values of \( A_3B_3 + B_3C_3 + C_3A_3 \).

3. **Upper Bound Calculation:**
   - We start with the given condition \( A_0B_0 + B_0C_0 + C_0A_0 = 1 \).
   - Since \( A_1B_1 = A_0B_0 \) and \( B_1C_1 = B_0C_0 \), we have:
     \[
     A_1B_1 + B_1C_1 = A_0B_0 + B_0C_0
     \]
   - Permuting \( A_1, B_1, C_1 \) to get \( A_2, B_2, C_2 \) does not change the sum of the distances:
     \[
     A_2B_2 + B_2C_2 + C_2A_2 = A_1B_1 + B_1C_1 + C_1A_1
     \]
   - Choosing \( A_3, B_3, C_3 \) such that \( A_3B_3 = A_2B_2 \) and \( B_3C_3 = B_2C_2 \), we get:
     \[
     A_3B_3 + B_3C_3 + C_3A_3 \leq 2(A_3B_3 + B_3C_3) = 2(A_2B_2 + B_2C_2)
     \]
   - Since \( A_2B_2 + B_2C_2 + C_2A_2 = A_1B_1 + B_1C_1 + C_1A_1 \), we have:
     \[
     A_3B_3 + B_3C_3 + C_3A_3 \leq 2(A_1B_1 + B_1C_1 + C_1A_1 - \min\{A_1B_1, B_1C_1, C_1A_1\})
     \]
   - Given \( A_1B_1 = A_0B_0 \) and \( B_1C_1 = B_0C_0 \), we can write:
     \[
     A_3B_3 + B_3C_3 + C_3A_3 \leq 2(2A_0B_0 + 2B_0C_0 - \min\{A_0B_0, B_0C_0\})
     \]
   - Simplifying further:
     \[
     A_3B_3 + B_3C_3 + C_3A_3 \leq 3(A_0B_0 + B_0C_0) + (A_0B_0 + B_0C_0 - 2\min\{A_0B_0, B_0C_0\})
     \]
   - Since \( A_0B_0 + B_0C_0 + C_0A_0 = 1 \):
     \[
     A_3B_3 + B_3C_3 + C_3A_3 \leq 3 \times 1 = 3
     \]

4. **Achieving the Upper Bound:**
   - Equality can be reached if \( A_0 = C_0 \), and the points \( A_1, B_1, C_1 \) are in this order on a straight line, and similarly for \( A_3, B_3, C_3 \).

5. **Lower Bound Calculation:**
   - By symmetry and similar reasoning, switching \( A_0, B_0, C_0, A_1, B_1, C_1 \) with \( A_3, B_3, C_3, A_2, B_2, C_2 \), we get that the minimum is \( \frac{1}{3} \).

The final answer is \( \boxed{\frac{1}{3} \text{ and } 3} \)