G11 (3-6, Romania) Given a plane $E$ and three non-collinear points $A, B, C$ on the same side of $E$, and the plane through $A, B, C$ is not parallel to plane $E$. Take any three points $A^{\prime}, B^{\prime}, C^{\prime}$ on plane $E$. Points $L, M, N$ are the midpoints of segments $A A^{\prime}, B B^{\prime}, C C^{\prime}$, respectively, and $O$ is the centroid of $\triangle L M N$ (excluding the case where the three points do not form a triangle). Find the locus of $O$ when $A^{\prime}, B^{\prime}$, $C^{\prime}$ vary arbitrarily on plane $E$.
Place all these points in the space $x y z$ coordinate system, and let $E$ be $z=0, A, B, C$ then in the upper half space $z>0$.

It is easy to know that the midpoint of two points $X=\left(x_{1}, x_{2}, x_{3}\right)$ and $Y=\left(y_{1}, y_{2}, y_{3}\right)$ is $\frac{X+Y}{2}=$ $\left(\frac{x_{1}+y_{1}}{2}, \frac{x_{2}+y_{2}}{2}, \frac{x_{3}+y_{3}}{2}\right)$, and the centroid of $\triangle X Y Z$ is $\frac{X+Y+Z}{3}$.

Therefore, the midpoints of segments $A A^{\prime}, B B^{\prime}, C C^{\prime}$ satisfy: $L=\frac{1}{2}\left(A+A^{\prime}\right), M=\frac{1}{2}(B$ $\left.+B^{\prime}\right), N=\frac{1}{2}\left(C+C^{\prime}\right)$, thus the centroid $O$ of $\triangle L M N$ satisfies
$$
\begin{aligned}
O & =\frac{1}{3}(L+M+N) \\
& =\frac{1}{6}\left(A+A^{\prime}+B+B^{\prime}+C+C^{\prime}\right) \\
& =\frac{1}{2}\left[\frac{1}{3}(A+B+C)+\frac{1}{3}\left(A^{\prime}+B^{\prime}+C^{\prime}\right)\right] \\
& =\frac{1}{2}\left(G+G^{\prime}\right) .
\end{aligned}
$$

where $G, G^{\prime}$ are the centroids of $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$, respectively. It is easy to see that $O$ lies in the plane $z=\frac{1}{6}\left(a_{3}\right.$ $+b_{3}+c_{3}$ ) (where $a_{3}, b_{3}, c_{3}$ are the $z$-coordinates of $A, B, C$), this plane is parallel to $z=0$ and the distance from $z=0$ is half the distance from $G$ to $z=0$.

Conversely, take any point $O$ in the plane $z=\frac{1}{6}\left(a_{3}+b_{3}+c_{3}\right)$, extend $G O$ to intersect $z=0$ at $G^{\prime}$, find three arbitrary points $A^{\prime}, B^{\prime}, C^{\prime}$ on $z=0$, such that $G^{\prime}$ is the centroid of $\triangle A^{\prime} B^{\prime} C^{\prime}$, and let the midpoints of $A A^{\prime}, B B^{\prime}, C C^{\prime}$ be $L, M, N$, respectively. From the previous calculation, we know: $O$ is the centroid of $\triangle L M N$.

Thus, the locus is the plane as described above, satisfying: it is parallel to $E$, the distance to $E$ is half the distance from the centroid $G$ of $\triangle A B C$ to $E$, and it is on the same side of $E$ as $G$.