6. On a two-way railway, there are 11 stations, each adjacent station being 7 kilometers apart. Starting at 7:00 AM, 18 freight trains are dispatched from the eleventh station in sequence, with one train every 5 minutes, all heading towards the first station at a speed of 60 kilometers per hour. At 8:00 AM, a passenger train is dispatched from the first station, heading towards the eleventh station at a speed of 100 kilometers per hour. Before reaching the terminal station, neither the freight trains nor the passenger train stop at any station. Question: Between which two adjacent stations can the passenger train meet 3 freight trains in succession?
6. Between the fifth and sixth stations, the passenger train meets 3 freight trains.
6.【Solution】A freight train is dispatched every 5 minutes, with a speed of 60 kilometers per hour, which is 1 kilometer per minute. Therefore, the distance between two consecutive freight trains is 5 kilometers.

The first freight train has been traveling for 1 hour before the passenger train departs, so the distance between the two trains is $7 \times (11-1) - 60 \times 1 = 10$ (kilometers),

The two trains meet after $\frac{10}{100+60} = \frac{1}{16}$ (hours),
meeting at a distance of $\frac{1}{16} \times 100 = \frac{25}{4}$ (kilometers) from the first station.
Since the distance between two consecutive freight trains is 5 kilometers, after the passenger train meets one freight train, it travels another $\frac{5}{100+60} \times 100 = \frac{25}{8}$ (kilometers),

before meeting the next freight train.
If A and B are two adjacent stations, then when the passenger train meets 3 freight trains between these two stations, the distance from the point where it meets the first freight train to point A should not exceed $7 - \frac{25}{8} \times 2 = \frac{3}{4}$ (kilometers).

Conversely, when this condition is met, the passenger train meets three freight trains between A and B. Suppose the passenger train meets the $(n+1)$-th freight train between stations A and B, and encounters two more freight trains between these two stations, then the passenger train has traveled $\frac{25}{4} + \frac{25}{8} n$ (kilometers) and the distance from the $(m+1)$-th station A is no more than $\frac{3}{4}$ kilometers, thus $\frac{25}{4} + \frac{25}{8} n - 7m \leq \frac{3}{4}$, which simplifies to $25(n+2) - 56m \leq 6$ (1)
Equation (1) indicates that a multiple of 25, when divided by 56, has a remainder $\leq 6$.

It is not difficult to verify that $25 \times 9 = 225 = 56 \times 4 + 1$, so the passenger train meets three freight trains between the 5th and 6th stations.

The next value that satisfies (1) is $25 \times 9 \times 2 = 56 \times 4 \times 2 + 2$, but at this point, $n+1 = 9 \times 2 - 1 = 17$. After the passenger train meets the $(n+1)$-th freight train, it can only meet one more freight train.

Therefore, the answer to this problem is: Between the 5th and 6th stations, the passenger train meets three freight trains. 【Note】If there are 19 or more freight trains, then between the 9th and 10th stations, the passenger train also meets three freight trains.