6. There are three points $A$, $B$, and $C$ in sequence on a straight line. A hunting dog at point $A$ discovers three rabbits, one large and two small, moving from point $B$ towards their burrow (point $C$), and immediately chases after them. When the rabbits notice the hunting dog, they hurry towards the burrow. The large rabbit can carry one small rabbit to increase its speed (the speed remains unchanged, and the time taken to pick up and put down the small rabbit is negligible). It is known that
$$
A B=a \mathrm{~m}, B C=b \mathrm{~m}\left(\frac{17}{27} b<a<3 b\right),
$$

The speeds of the hunting dog, the large rabbit, and the small rabbit are $2 \mathrm{~m} / \mathrm{s}$, $1.5 \mathrm{~m} / \mathrm{s}$, and $1 \mathrm{~m} / \mathrm{s}$, respectively, and the rabbits' initial speed is $0.5 \mathrm{~m} / \mathrm{s}$. The three rabbits can at most be $\qquad$ $\mathrm{m}$ away from point $B$ when they notice the hunting dog, to just make it into the burrow before the hunting dog catches them.
6. $\frac{27 a-17 b}{64}$.

Obviously, the hound needs $\frac{a+b}{2} \mathrm{~s}$ to reach the rabbit's den.
Assume the rabbit is $x \mathrm{~m}$ away from point $B$ when it spots the hound, having already used $2 x \mathrm{~s}$. Then, for the remaining $(b-x) \mathrm{m}$, the large rabbit must lead the small rabbit to finish the distance within $\left(\frac{a+b}{2}-2 x\right) \mathrm{s}$. We now find $x_{\text {max }}$.
Consider a segment $D E$ of length $y \mathrm{~m}$.
Assume the three rabbits need at least $t \mathrm{~s}$ to complete the distance. The large rabbit must first carry one small rabbit to a point $G$ ahead, then let this small rabbit continue running on its own, while the large rabbit runs back to meet the other small rabbit at point $F$, and then carry it forward, eventually arriving at point $E$ simultaneously with the first small rabbit.

Since the speed ratio of the large rabbit to the small rabbit is $3: 2$, the entire distance should be divided into three segments in the ratio
$D F: F G: G E=4: 1: 4$.
Thus, $t=\frac{11 y}{9} \div \frac{3}{2}=\frac{22 y}{27}$.
It is easy to see that dividing the entire distance into several segments and running each segment in the above manner takes the same amount of time as running the entire distance in the above manner. Therefore,
$$
\begin{array}{l}
\frac{a+b}{2}-2 x \geqslant \frac{22}{27}(b-x) \\
\Rightarrow x \leqslant \frac{27 a-17 b}{64} \Rightarrow x_{\max }=\frac{27 a-17 b}{64} .
\end{array}
$$