If x satisfies |3 x-2|+|3 x-4|+|3 x-6| ≥ 12 , then

Dividing $\mathrm{R}$ at $\frac{2}{3}, \frac{4}{3}$ and $2,$ analyse 4 cases. When $\mathrm{x} \leq \frac{2}{3},$ the inequality becomes $2-3 x+4-3 x+6-3 x \geq 12$ implying $-9 x \geq 0 \Rightarrow x \leq 0$ when $\mathrm{x} \geq 2$ the ineqality becomes $3 x-2+3 x-4+3 x-6 \geq 12$ Implying $9 \mathrm{x} \geq 24 \Rightarrow \mathrm{x} \geq 8 / 3$ The inequality in invalid in the other two sections. $\therefore$ either $x \leq 0$ or $x \geq 8 / 3$