Given that on the same side of plane $E$ there are three non-collinear points $A, B, C$, and the plane passing through $A, B, C$ is not parallel to plane $E$. Take any three points $A^{\prime}, B^{\prime}, C^{\prime}$ on plane $E$, and let points $L, M, N$ be the midpoints of line segments $A A^{\prime}, B B^{\prime}, C C^{\prime}$, respectively, and $L, M, N$ are not collinear. Let $G$ be the centroid of $\triangle L M N$. When $A^{\prime}, B^{\prime}, C^{\prime}$ are arbitrarily chosen on plane $E$, find the locus of point $G$.

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Parse the centroid $G$ of $\triangle L M N$ and the centroids $P$ and $P^{\prime}$ of $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$. What is the relationship between them? As shown in the figure, let the midpoints of segments $A B$, $L M$, and $A^{\prime} B^{\prime}$ be $D$, $H$, and $D^{\prime}$, respectively. Intuition tells us: $D, H$, and $D^{\prime}$ are collinear and $H$ is the midpoint of $D D^{\prime}$. Is this really the case? In fact, we have the following conclusion:

Connecting the midpoints $E, F, G, H$ of the sides of a spatial quadrilateral $A B C D$ forms a parallelogram.

As shown in Figure b, connect $A C$ and $B D$. By $E F \perp \frac{1}{2} A C \perp H G, E H \perp \frac{1}{2} B D \perp F G$, and $A C \times B D$, the above conclusion holds.
Figure a
Figure b
This proves that $H$ is the midpoint of line $D D^{\prime}$. This leads us to conjecture: $P, G, P^{\prime}$ are collinear and $G$ is the midpoint of segment $P P^{\prime}$. To prove this conjecture, let the other trisection points of the medians $C D$ and $C^{\prime} D^{\prime}$, except for the centroid, be $Q$ and $Q^{\prime}$, respectively. As shown in Figure a, consider the quadrilateral $D D^{\prime} Q^{\prime} Q$. The line segments connecting the midpoints of the opposite sides, $P P^{\prime}$ and $H K$, bisect each other at $G^{\prime}$. Now consider the quadrilateral $C C^{\prime} P^{\prime} P$. The line segments connecting the midpoints of the opposite sides, $N G^{\prime}$ and $Q Q^{\prime}$, bisect each other at $K^{\prime}$. Since $K$ and $K^{\prime}$ are both midpoints of $Q Q^{\prime}$, $K=K^{\prime}$. Therefore, $N, K, G^{\prime}, H$ are collinear, and $K$ and $G^{\prime}$ are the trisection points of segment $N H$. Thus, $G^{\prime}$ is the centroid of $\triangle L M N$, and $G^{\prime}=G$.

Draw a perpendicular from the centroid $P$ of $\triangle A B C$ to plane $E$ at $T$, and let the midpoint of $P T$ be $S$. The plane parallel to plane $E$ and passing through $S$ is $\alpha$. It is known that point $G$ lies in plane $\alpha$.

Next, we prove the purity of the trajectory. Let $G$ be any point in plane $\alpha$, and extend $P G$ to intersect plane $E$ at $P^{\prime}$. As shown in Figure c, it is easy to see that $G$ is the midpoint of $P P^{\prime}$. Take any two points $A^{\prime}$ and $B^{\prime}$ in plane $E$ that are not collinear with $P^{\prime}$. Let $D^{\prime}$ be the midpoint of $A^{\prime} B^{\prime}$. On the extension of $D^{\prime} P^{\prime}$, take point $C^{\prime}$ such that $C^{\prime} D^{\prime} = 3 P^{\prime} D^{\prime}$. Connect $A A^{\prime}$, $B B^{\prime}$, and $C C^{\prime}$, and let the midpoints of $A A^{\prime}$, $B B^{\prime}$, and $C C^{\prime}$ be $L$, $M$, and $N$, respectively. As before, it can be shown that $G$ is the centroid of $\triangle L M N$.
Figure c
Figure d
Parse 2 As shown in Figure d, take plane $E$ as the coordinate plane $z=0$, and any point in plane $E$ as the origin $O$, to establish a spatial rectangular coordinate system $O-x y z$. Let the coordinates be $A\left(x_{1}, y_{1}, 2 a\right)$, $B\left(x_{2}, y_{2}, 2 b\right)$, $C\left(x_{3}, y_{3}, 2 c\right)$; $A^{\prime}\left(x_{1}^{\prime}, y_{1}^{\prime}, 0\right)$, $B^{\prime}\left(x_{2}^{\prime}, y_{2}^{\prime}, 0\right)$, $C^{\prime}\left(x_{3}^{\prime}, y_{3}^{\prime}, 0\right)$. Then the coordinates of the midpoints of $A A^{\prime}$, $B B^{\prime}$, and $C C^{\prime}$ are $L\left(\frac{x_{1}+x_{1}^{\prime}}{2}, \frac{y_{1}+y_{1}^{\prime}}{2}, a\right)$, $M\left(\frac{x_{2}+x_{2}^{\prime}}{2}, \frac{y_{2}+y_{2}^{\prime}}{2}, b\right)$, $N\left(\frac{x_{3}+x_{3}^{\prime}}{2}, \frac{y_{3}+y_{3}^{\prime}}{2}, c\right)$. The centroid $G$ of $\triangle L M N$ has coordinates
$$
G\left(\frac{x_{1}+x_{2}+x_{3}+x_{1}^{\prime}+x_{2}^{\prime}+x_{3}^{\prime}}{6}, \frac{y_{1}+y_{2}+y_{3}+y_{1}^{\prime}+y_{2}^{\prime}+y_{3}^{\prime}}{6}, \frac{a+b+c}{3}\right).
$$

It is evident that, regardless of the positions of $A^{\prime}$, $B^{\prime}$, and $C^{\prime}$ in plane $E$, the $z$-coordinate of point $G$ is always a constant $\frac{a+b+c}{3}$. That is, point $G$ is always on a plane parallel to plane $E$, $z=\frac{a+b+c}{3}$. On the other hand, since $A^{\prime}$, $B^{\prime}$, and $C^{\prime}$ can move arbitrarily in plane $E$, $x_{i}^{\prime}$ and $y_{i}^{\prime}$ ($i=1,2,3$) can take any real values. Therefore, the $x$ and $y$ coordinates of point $G$
$$
x_{G}=\frac{x_{1}+x_{2}+x_{3}+x_{1}^{\prime}+x_{2}^{\prime}+x_{3}^{\prime}}{6}, y_{G}=\frac{y_{1}+y_{2}+y_{3}+y_{1}^{\prime}+y_{2}^{\prime}+y_{3}^{\prime}}{6}
$$

can take any real values. Thus, the moving point $G$ can cover all points on the plane $z=\frac{a+b+c}{3}$. Therefore, the trajectory of the moving point $G$ is the plane $z=\frac{a+b+c}{3}$.