One. (20 points) Two cars, A and B, start from point A at the same time, driving in the same direction on a straight line. Car A can carry at most $a \mathrm{~L}$ of gasoline, and Car B can carry at most $b \mathrm{~L}$ of gasoline $(a \geqslant b$ and both are the maximum capacities of the fuel tanks). They cannot refuel during the journey, but the two cars can lend each other fuel. Ultimately, both cars must return to point A. Design a scheme to make one of the cars travel as far away from the starting point A as possible, and calculate how many kilometers this car has traveled in total? (Both cars have the same fuel consumption rate, and 1 liter of fuel can make a car travel $12 \mathrm{~km}$.)
One car travels from $A$ to $B$, and another car travels from $A$ to $B$. Car B stays at $B$, first lending some gasoline to Car A to fill its tank. Car A travels the farthest to $C$ using half of the gasoline in the tank, and then returns to $B$ using the other half of the gasoline in the tank. At this point, Car B lends gasoline to Car A for the second time, and both cars return to $A$ together.

Let the gasoline used by Car A from $A$ to $C$ be $x \mathrm{~L}$, and the gasoline used by Car B from $A$ to $B$ be $y$ L. Then
$$
\left\{\begin{array}{l}
x+y=\frac{1}{2}(a+b), \\
x-y=\frac{1}{2} a .
\end{array}\right.
$$

Solving these equations, we get $x=\frac{2 a+b}{4}, y=\frac{b}{4}$.
Therefore, the total distance traveled by this car (Car A) is
$$
12 \times \frac{2 a+b}{4} \times 2=6(2 a+b)(\mathrm{km}) .
$$