A bag contans 2017 red balls and 2017 black balls. We remove two balls at a time repeatedly and (1) discard them if they are of some colour (2) discard the black ball and return to the bag the red ball if they are of different colours. Then the probability that this process will terminate with one red ball is `'p'` where `3p` is______

Explanation

To solve the problem, we need to analyze the process of removing balls from the bag and the implications of the rules given.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
We start with 2017 red balls and 2017 black balls in the bag.

2. **Analyzing the Removal Process**:
- If we pick two balls of the same color (either both red or both black), we discard both.
- If we pick one red ball and one black ball, we discard the black ball and return the red ball to the bag.

3. **Impact of the Rules**:
- The first case (removing two balls of the same color) reduces the total number of balls by 2.
- The second case (removing one red and one black) reduces the total number of balls by 1 (since we discard the black ball and keep the red ball).

4. **Observing the Color Counts**:
- The number of red balls can only stay the same or decrease.
- The number of black balls can only decrease.

5. **Final Outcome**:
- The process continues until there are no black balls left. At that point, only red balls will remain.
- Since we started with an equal number of red and black balls, and we can only remove black balls, we will eventually be left with only red balls.

6. **Probability Calculation**:
- The process will always terminate with at least one red ball remaining.
- Therefore, the probability \( p \) that the process will terminate with one red ball is \( p = 1 \).

7. **Finding \( 3p \)**:
- Since \( p = 1 \), we calculate \( 3p = 3 \times 1 = 3 \).

### Final Answer:
Thus, the value of \( 3p \) is **3**.