3. (10 points) Car $A$ departs from station A heading to station B, while cars $B$ and $C$ depart from station B heading towards station A at the same time. On the way, $A$ meets $B$ 20 minutes after meeting $C$. It is known that the speeds of $A$, $B$, and $C$ are 90 km/h, 80 km/h, and 60 km/h, respectively. The distance between station A and station B is $\qquad$ km.
【Analysis】According to the problem, A meets B 20 minutes before meeting C, from which we can find that A and C travel 50 kilometers in 20 minutes (1/3 hour): $(90+60) \times \frac{1}{3}=50$ kilometers. These 50 kilometers are the extra distance B travels compared to C during the time A and B meet. Clearly, the time A and B meet is $50 \div (80-60)=2.5$ hours. Then, based on the sum of speeds $\times$ meeting time $=$ distance between the two places, we can solve the problem by setting up the equation.
【Solution】Solution: 20 minutes $=\frac{1}{3}$ hour,
A and C meet after 20 minutes, traveling a total of $(90+60) \times \frac{1}{3}=50$ (kilometers),
These 50 kilometers are the extra distance B travels compared to C during the time A and B meet. Clearly, the time A and B meet is $50 \div (80-60)=2.5$ (hours).

Therefore, the distance between stations A and B when A and B meet is $(90+80) \times 2.5=425$ (kilometers).
Answer: The distance between stations A and B is 425 kilometers.
So the answer is: 425.