6. In the Cartesian coordinate system, the area of the plane region corresponding to the point set $\{(x, y) \mid(|x|+|3 y|-6)(|3 x|+|y|-6) \leq 0\}$ is $\qquad$ .


Answer: 24. Solution: Let $K_{1}=\{(x, y) \| x|+| 3 y \mid-6 \leq 0\}$. First, consider the part of $K_{1}$ in the first quadrant, where $x+3 y \leq 6$. Thus, these points correspond to the region of $\triangle \mathrm{OCD}$ and its interior in the graph. By symmetry, the region corresponding to $K_{1}$ is the rhombus $\mathrm{ABCD}$ and its interior, centered at the origin $\mathrm{O}$.

Similarly, let $K_{2}=\{(x, y) \| 3 x|+| y \mid-6 \leq 0\}$, then the region corresponding to $K_{2}$ is the rhombus $\mathrm{EFGH}$ and its interior, centered at $\mathrm{O}$.

By the definition of the point set $K$, the plane region corresponding to $K$ is the part covered by exactly one of $K_{1}$ or $K_{2}$. Therefore, what is required in this problem is the area $S$ of the shaded region in the graph.

Since the equation of line $\mathrm{CD}$ is $x+3 y=6$, and the equation of line $\mathrm{GH}$ is $3 x+y=6$, the coordinates of their intersection point $\mathrm{P}$ are $\left(\frac{3}{2}, \frac{3}{2}\right)$. By symmetry, $S=8 S_{\triangle C P G}=8 \times \frac{1}{2} \times 4 \times \frac{3}{2}=24$.