6. In the Cartesian coordinate system $x O y$, the area of the plane region corresponding to the point set $K=\{(x, y) \mid(|x|+|3 y|-6)(|3 x|+|y|-6) \leqslant 0\}$ is $\qquad$ .
The figure corresponding to the point set $K$ is shown in the shaded area on the right.
The intersection point coordinates of the two lines $x+3 y-6=0$ and $3 x+y-6=0$ located in the first quadrant are $A\left(\frac{3}{2}, \frac{3}{2}\right)$. Thus, the area of each small triangle is $\frac{1}{2} \times 4 \times \frac{3}{2}=3$, so the area of the shaded part is $3 \times 8=24$.