Example 4. As shown in Figure 6, the dihedral angle N-PQ-M is a $120^{\circ}$ dihedral angle. From a point $\mathrm{A}$ inside it, perpendicular lines $A B$ and $A C$ are drawn to the planes $\mathrm{M}$ and $\mathrm{N}$, respectively. If $A B=3, A C=1$, find the distance from point $A$ to the edge $P Q$.
Analyzing this problem, if we do not approach it by constructing an auxiliary plane, but instead draw $AD \perp PQ$ through $\mathbf{A}$, and assume that $AC$, $AD$, and $AB$ are not in the same plane, it would be incorrect, and the problem would be unsolvable.

Let's look at the conditions given in the problem. Since $\mathrm{AC} \perp \mathbf{N}$ and $\mathrm{AB} \perp \mathrm{M}$, we have $\mathbf{AC} \perp PQ$ and $\mathrm{AB} \perp PQ$. Therefore, $PQ \perp \mathbf{AC}$ and $PQ \perp \mathbf{AB}$, which means that $PQ$ is perpendicular to the plane defined by $\mathbf{AC}$ and $\mathbf{AB}$. This implies that an auxiliary plane can be constructed through $\mathbf{AC}$ and $\mathbf{AB}$, and $AD \perp PQ$, making $AD$ the distance from $A$ to $PQ$.

Solution: Construct a plane through $\mathrm{AB}$ and $\mathrm{AC}$ intersecting $\mathrm{PQ}$ at $\mathrm{D}$, and intersecting planes $M$ and $N$ at $BD$ and $CD$ respectively. Since $A B \perp M$ and $A C \perp N$, we have $A B \perp P Q$ and $A C \perp P Q$, meaning $P Q$ is perpendicular to the plane through $A B$ and $A C$.
Thus, $P Q \perp C D$ and $P Q \perp B D$. Therefore, $\angle C D B = 120^{\circ}$.
Also, since $\mathrm{AC} \perp \mathrm{CD}$ and $\mathrm{AB} \perp \mathrm{DB}$,
$\therefore A, C, D, B$ are concyclic, with $AD$ as the diameter of the circle, and $\angle \mathrm{CAB} = 60^{\circ}$.
Connecting $C \mathbf{B}$, in $\triangle \mathrm{A} C \mathbf{B}$, by the cosine rule:
\[
\begin{aligned}
\mathrm{CB}^{2} & =\mathbf{A B}^{2}+\mathbf{A} \mathrm{C}^{2}-2 \mathbf{A B} \cdot \mathbf{A C} \cdot \cos \angle \mathrm{CAB} \\
& =3^{2}+1^{2}-2 \times 3 \times \cos 60^{\circ}=7 .
\end{aligned}
\]

Thus, $\mathrm{CB} = \sqrt{7}$.
By the sine rule: $\mathrm{BC} = 2 \mathrm{R} \sin \mathrm{A} = \mathrm{AD} \cdot \sin 60^{\circ}$, so $\mathrm{AD} = \frac{\mathrm{B} C}{\sin 60^{\circ}} = \frac{\sqrt{7} \cdot 3}{2} = \frac{2 \sqrt{21}}{3}$.