One, (This question is worth 20 points)
In the border desert area, patrol vehicles travel 200 kilometers every day. Each vehicle departs from base $A$ at the same time, completes the mission, and then returns along the original route to the base (returning together). After vehicles A and B reach point $B$ on the way, they only keep enough gasoline for their own return to the base, and leave the extra gasoline for the other three vehicles to use. How far can the other three vehicles travel at most in kilometers?
When the car reaches point $B$ on the way, it takes $x$ days, and from $B$ to the farthest point, it takes $y$ days, then we have
$$
2[3(x+y)+2 x]=14 \times 5,
$$

which simplifies to $5 x+3 y=35$.
According to the problem, we need $x>0, y>0$, and
$$
14 \times 5-(5+2) x \leqslant 14 \times 3 \text {, }
$$

which simplifies to $x \geqslant 4$.
Thus, the essence of the problem is to find the maximum value of $y$ under the constraints
$$
\left\{\begin{array}{l}
5 x+3 y=35, \\
x \geqslant 4, \\
y>0
\end{array}\right.
$$

Clearly, $y=5$.
Therefore, $200 \times(4+5)=1800$ (km) is the farthest distance the other three cars can travel.