11. If the distances from the three vertices of a regular $\triangle A B C$ with side length 6 to the plane $\alpha$ are $1,2,3$, then the distance from the centroid $G$ of $\triangle A B C$ to the plane $\alpha$ is $\qquad$ .
Solve $\left\{0, \frac{2}{3}, \frac{4}{3}, 2\right\}$.
Analysis 1) When the three vertices of $\triangle A B C$ are on the same side of plane $\alpha$, the distance from the centroid $G$ to plane $\alpha$ is calculated using the formula $d=\frac{h_{1}+h_{2}+h_{3}}{3}$, which gives 2;
2) When one of the three vertices of $\triangle A B C$ is on the opposite side of plane $\alpha$ compared to the other two, the distances from the centroid $G$ to plane $\alpha$ are calculated as $0, \frac{2}{3}, \frac{4}{3}$.