Tourists planned a long three-day tour, with the intention of walking a third of the entire route each day. They only kept this plan on the first day. On the second day, they only walked a third of the remaining distance, and on the third day, exhausted, they only managed a quarter of the remaining distance. The last $24 \mathrm{~km}$ to the destination they were driven by an off-road vehicle.

How long was the entire tour supposed to be, and how many kilometers did the tourists walk on the first, second, and third day?
On the first day, the tourists walked $\frac{1}{3}$ of the journey, leaving $\frac{2}{3}$ of the total distance to the destination.

On the second day, they walked $\frac{1}{3}$ of the remaining distance from the previous day, which is $\frac{1}{3} \cdot \frac{2}{3}=\frac{2}{9}$ of the entire journey; the remaining distance to the destination was $\frac{2}{3}-\frac{2}{9}=\frac{4}{9}$ of the entire journey.

Similarly, on the third day, they walked $\frac{1}{4}$ of $\frac{4}{9}$, which is $\frac{1}{9}$ of the entire route. At this point, they had $\frac{4}{9}-\frac{1}{9}=\frac{1}{3}$ of the total planned route left, which, according to the problem, was $24 \mathrm{~km}$. Therefore, the entire trip was $3 \cdot 24=72(\mathrm{~km})$ long.

From this, we can easily calculate the distance the tourists walked each day:

- on the first day, it was $\frac{1}{3} \cdot 72=24(\mathrm{~km})$,
- on the second day, $\frac{2}{9} \cdot 72=16(\mathrm{~km})$,
- on the third day, $\frac{1}{9} \cdot 72=8(\mathrm{~km})$.