Three. (20 points) As shown in Figure 5, there are two rectangular metal parts $ABCD$ and $ECFG$ placed on the Cartesian coordinate plane. A laser beam emitted from the origin $O$ will cut the parts. If the laser $(l)$ cuts two metal parts (both above $l$) from the two rectangles with exactly equal areas in one go, find the equation of the line $(l)$.

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Three. (20 points) As shown in Figure 5, there are two rectangular metal parts $ABCD$ and $ECFG$ placed on the Cartesian coordinate plane. A laser beam emitted from the origin $O$ will cut the parts. If the laser $(l)$ cuts two metal parts (both above $l$) from the two rectangles with exactly equal areas in one go, find the equation of the line $(l)$.
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Three, it is known that $k_{O C}=\frac{2}{5}, k_{O E}=\frac{4}{7}, k_{O F}=\frac{3}{5}, k_{O G}=\frac{6}{7}$.
Let the equation of line $l$ be $y=k x$.
(1) When $k_{O F} \leqslant k \leqslant k_{O G}$, i.e., $\frac{3}{5} \leqslant k \leqslant \frac{6}{7}$, let $l$ intersect the boundaries of the two metal parts at points $A^{\prime}, C^{\prime}, G^{\prime}, F^{\prime}$ (as shown in Figure 8). It is easy to know that $A^{\prime}(4,4 k), G^{\prime}\left(\frac{4}{k}, 4\right), G^{\prime}(7,7 k)$, $F^{\prime}\left(\frac{6}{k}, 6\right)$.
Since $S_{\triangle X^{\prime} C D}=S_{\triangle G F G}$, and
Rt $\triangle A^{\prime} C^{\prime} D \backsim$ Rt $\triangle G^{\prime} F^{\prime} G$,
thus Rt $\triangle A^{\prime} C^{\prime} D \cong$ Rt $\triangle G^{\prime} F^{\prime} G$.
By $D C=G F^{\prime}$, we get $\frac{4}{k}-4=\frac{6}{k}-7$. Therefore, $k=\frac{2}{3} \in\left[\frac{3}{5}, \frac{6}{7}\right)$.
(2) When $k_{O E} \leqslant k<k_{O F}$, i.e., $\frac{4}{7} \leqslant k<\frac{3}{5}$, let $l$ intersect the boundaries of the two metal parts at points $A^{\prime \prime}, C^{\prime \prime}, G^{\prime \prime}, B^{\prime}$ (as shown in Figure 8). It is easy to know that $A^{\prime \prime}(4,4 k), C^{\prime \prime}\left(\frac{4}{k}, 4\right), G^{\prime \prime}(7,7 k)$, $B^{\prime}(10,10 k)$.
By $S_{\triangle r C D}=S_{\text {quadrilateral } B}$, we get
$$
\frac{1}{2}\left(\frac{4}{k}-4\right)(4-4 k)=\frac{1}{2}[(6-7 k)+(6-10 k)] \cdot 3 \text {, }
$$

which simplifies to $67 k^{2}-68 k+16=0$.
Solving, we get $k_{1,2}=\frac{2(17 \pm \sqrt{21})}{67}$
$\approx 0.644256$ or 0.370669.
It is easy to see that $k_{1,2} \notin\left[\frac{4}{7}, \frac{3}{5}\right)$. This is not possible.
(3) When $k_{O C}<k<k_{O E}$, i.e., $\frac{2}{5}<k<\frac{4}{7}$, let $l$ intersect the boundaries of the two metal parts at points $A^{\prime \prime \prime}, C^{\prime \prime}, B^{\prime \prime}$ (as shown in Figure 8). It is easy to know that $A^{\prime \prime \prime}(4,4 k), C^{\prime \prime}\left(\frac{4}{k}, 4\right), B^{\prime \prime}(10,10 k)$.
By $S_{\triangle r C D}=S_{\text {pentagon } E C B r C C}$, we get
$S_{\triangle T C D}=S_{\text {quadrilateral } E C F C}-S_{\triangle B C C C}$.
Thus, $\frac{1}{2}\left(\frac{4}{k}-4\right)(4-4 k)$
$$
=6-\frac{1}{2}\left(10-\frac{4}{k}\right)(10 k-4) \text {. }
$$

which simplifies to $29 k^{2}-31 k+8=0$.
Solving, we get $k_{1,2}=\frac{31 \pm \sqrt{33}}{58} \approx 0.6335$ or 0.435.
Only $k=\frac{31-\sqrt{33}}{58} \in\left(\frac{2}{5}, \frac{4}{7}\right)$.
In summary, the desired equations of the line are
$y=\frac{2}{3} x$ or $y=\frac{31-\sqrt{33}}{58} x$.
(Gu Huda provided)