According to the conditions, we can get MX = Y+ZNY = X+ZKZ = X+Y.

With these three terms, (M+ 1)X=(N+ 1)Y and (K+ 1)Z=(N+ 1)Y can be derived.

According to the topic, we can get MX/(n+1) y+ny/(n+1) y+kz/(n+1) y.

After substituting the first formula and merging it into a fraction, we can get: 2 (x+y+z)/(ny+y) = 2 (x+y+z)/(x+y+z) = 2.

I don't know if this idea is correct.

Just for reference. I imagine that my father will send his youngest son for a while first, then let him go, pick up his eldest son, and send him directly to his grandmother's house, arriving at the same time as his youngest son. I think this should be the fastest way, because the distance is the shortest.

Now suppose that the time to send the youngest son is X, the time to pick up the eldest son is Y, and the time to send the eldest son is Z:

According to the topic, the condition set is: 1)20X+5Y+5Z=33 (for the younger son) 2)5X+5Y+20Z=33 (for the eldest son) 3)20X+5Y+20Z=33.

There are three sets of conditions (taking the father as the reference object) that can be obtained as 25X+5Y=33 and x = y.

Suppose that if we can arrive within 3 hours, then X+Y+Z is less than or equal to 3, which means that 2X+Y is equal to 3 at most. Then, according to the conditions, y can be less than or equal to 0.6 X, and greater than or equal to 1. 1, both of which have positive solutions, so it is possible.