In the plane rectangular coordinate system $x O y$, the set of points
$$
\begin{aligned}
K= & \{(x, y) \mid(|x|+|3 y|-6) \cdot \\
& (|3 x|+|y|-6) \leqslant 0\}
\end{aligned}
$$

corresponds to a plane region whose area is $\qquad$
6. 24 .

Let $K_{1}=\{(x, y)|| x|+| 3 y |-6 \leqslant 0\}$.
First, consider the part of the point set $K_{1}$ in the first quadrant, at this time, $x+3 y \leqslant 6$. Therefore, these points correspond to $\triangle O C D$ and its interior in Figure 2.

By symmetry, the region corresponding to the point set $K_{1}$ is the rhombus $A B C D$ and its interior centered at the origin $O$ in Figure 2.
Similarly, let
$$
K_{2}=\{(x, y)|| 3 x|+| y |-6 \leqslant 0\} \text {. }
$$

Then the region corresponding to the point set $K_{2}$ is the rhombus $E F G H$ and its interior centered at $O$ in Figure 2.

By the definition of the point set $K$, the plane region corresponding to $K$ is the part covered by exactly one of the point sets $K_{1}$ and $K_{2}$.

Therefore, what is required in this problem is the area $S$ of the shaded region in Figure 2.

By $l_{c D}: x+3 y=6, l_{C H}: 3 x+y=6$, we know that the intersection point of the two lines is $P\left(\frac{3}{2}, \frac{3}{2}\right)$.
By symmetry,
$$
S=8 S_{\triangle C P G}=8 \times \frac{1}{2} \times 4 \times \frac{3}{2}=24 .
$$