5. In the border desert area, patrol vehicles travel 200 kilometers per day, and each vehicle can carry enough gasoline to travel for 14 days. There are 5 patrol vehicles that set out from base $A$ simultaneously, complete their tasks, and then return along the same route to the base. To allow three of the vehicles to patrol as far as possible (and then return together), vehicles 甲 and 乙 travel to a point $B$ along the way, leaving only enough gasoline for their return to the base, and giving the remaining gasoline to the other three vehicles. How far can the other three vehicles travel in kilometers?
5. Suppose the patrol car travels to point $B$ in $x$ days, and three of the cars take $y$ days to travel from $B$ to the farthest point, then we have
$$
2[3(x+y)+2 x]=14 \times 5,
$$

which simplifies to $5 x+3 y=35$.
From the problem, we know that $x>0, y>0$ and $14 \times 5-(5+2) x \leqslant 14 \times 3$, which means $x \geqslant 4$. From (1), we get
$$
y=\frac{35-5 x}{3} \text {. }
$$

To maximize the distance traveled, we need to find the maximum value of $y$, which occurs when $x$ is at its minimum value, i.e., $x=4$. Thus, $y=5$.
Therefore, $200 \times(4+5)=1800$ (km), meaning the farthest distance the other three cars can travel is 1800 km.