In the cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$ with edge length $a\left(a \in \mathbf{R}_{+}\right)$ as shown in Figure 4, $E$ is the center of the square $A B B_{1} A_{1}$, and $F, G$ are points moving along the edges $B_{1} C_{1}$ and $D D_{1}$, respectively. Question: What is the maximum value of the area of the projection of the spatial quadrilateral DEFG onto the six faces of the cube?
The projection area of the spatial quadrilateral $D E F G$ on the bottom surface is equal, and its orthogonal projection on the bottom surface $A B C D$ is the $\triangle D E_{1} F_{1}$ shown in Figure 5.

Since point $F_{1}$ slides along $B C$, the maximum area is $S_{\triangle D E_{1} C}=\frac{1}{2} a^{2}$, and the minimum area is $S_{\triangle D E_{1} n}=\frac{1}{4} a^{2}$.

The projection area of the spatial quadrilateral $D E F G$ on the left and right side surfaces is equal, and its orthogonal projection on the left side surface $D C C_{1} D_{1}$ is the $\triangle D C_{1} G$ shown in Figure 6, with the maximum area being $S_{\triangle D C_{1} D_{1}}=\frac{1}{2} a^{2}$, and the minimum area being 0 (at this time, point $G$ coincides with point $D$).

The projection area of the spatial quadrilateral $D E F G$ on the front and back side surfaces is equal, and its orthogonal projection on the front side surface $D A A_{1} D_{1}$ is the quadrilateral $D E_{3} F_{3} G$ shown in Figure 7, with the maximum area being $S_{\text {quadrilateral } D E B_{1} N_{1} D_{1}}=\frac{3}{4} a^{2}$, and the minimum area being $S_{\triangle D E_{3} A_{1}}=\frac{1}{4} a^{2}$.
In summary, the maximum value of the area of the formed figure is $\frac{3}{4} a^{2}$.