6. In a Cartesian coordinate system, the "traffic distance" between points $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$ is defined as
$$
d(P, Q)=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right| \text {. }
$$

A certain country has 3 castles, marked on the map as $A(2,3)$, $B(-6,9)$, and $C(-3,-8)$. The country plans to build a new castle. If the traffic distance from this new castle to the other 3 castles is equal, then the location of the new castle is $\qquad$
6. $(-5,0)$.

Let the new castle be $D(x, y)$.
If $x \geqslant 2$, as shown in Figure 2,
when $y \geqslant 3$, $d(A, D) < d(C, D)$; when $y < 3$,
$d(A, D) < d(B, D)$. This is a contradiction, so $x < 2$.
Similarly, we know
$-8 < y < 9$.
Thus, $d(A, D) = |x-2| + |y-3| = 2-x + |y-3|$,
$d(B, D) = |x+6| + |y-9| = |x+6| + 9-y$,
$d(C, D) = |x+3| + |y+8| = |x+3| + y + 8$.
By $d(B, D) = d(C, D)$, we get
$$
\begin{array}{l}
2 y = |x+6| - |x+3| + 1 \\
\leqslant |(x+6) - (x+3)| + 1 = 4 .
\end{array}
$$

Thus, $y \leqslant 2$.
Then, $d(A, D) = 2-x + (3-y) = 5-x-y$.
By $d(A, D) = d(B, D)$, we get
$$
\begin{array}{l}
5-x-y = |x+6| + 9-y \\
\Rightarrow |x+6| = -x-4 \Rightarrow x = -5 .
\end{array}
$$

Then, $y = \frac{1}{2}(|x+6| - |x+3| + 1) = 0$.
Therefore, $D(-5,0)$.