Example 6 Given $B P: P Q: Q C=1: 2: 1, C G: A G$ $=1: 2$. Then $B E: E F: F G=(\quad$.
(A) $12: 17: 7$
(B) 11:16:6
(C) $10: 15: 6$
(D)9:14:5
(15th Wuyang Cup Junior High School Mathematics Competition (Grade 9))
Solution: As shown in Figure 7, let
$$
\begin{array}{l}
\angle C B G=a, \angle G B A \\
=\beta, B P=Q C=x, \\
P Q=2 x, B A=y .
\end{array}
$$

From the perspective of point $B$,
applying the Angle Bisector Theorem to points $P, E, A$, $Q, F, A$, and $C, G, A$ respectively, we get
$$
\begin{array}{l}
\frac{\sin (\alpha+\beta)}{B E}=\frac{\sin \alpha}{y}+\frac{\sin \beta}{x}, \\
\frac{\sin (\alpha+\beta)}{B F}=\frac{\sin \alpha}{y}+\frac{\sin \beta}{3 x}, \\
\frac{\sin (\alpha+\beta)}{B G}=\frac{\sin \alpha}{y}+\frac{\sin \beta}{4 x} .
\end{array}
$$

From equations (1), (2), and (3), we have
$$
\begin{array}{l}
B E=\frac{x y \sin (\alpha+\beta)}{x \sin \alpha+y \sin \beta}, \\
B F=\frac{3 x y \sin (\alpha+\beta)}{3 x \sin \alpha+y \sin \beta}, \\
B C=\frac{4 x y \sin (\alpha+\beta)}{4 x \sin \alpha+y \sin \beta} .
\end{array}
$$

Also, from $C G: A G=1: 2$, we know
$$
\frac{S_{\triangle B C}}{S_{\triangle B C}}=\frac{\frac{1}{2} B G \cdot y \sin \beta}{\frac{1}{2} B G \cdot 4 x \sin \alpha}=\frac{y \sin \beta}{4 x \sin \alpha}=\frac{2}{1} \text {. }
$$

Thus, $y \sin \beta=8 x \sin \alpha$.
Substituting equation (7) into equations (1), (2), and (3), we get
$$
\begin{array}{l}
B E=\frac{y \sin (\alpha+\beta)}{9 \sin \alpha}, \\
B F=\frac{3 y \sin (\alpha+\beta)}{11 \sin \alpha}, \\
B G=\frac{y \sin (\alpha+\beta)}{3 \sin \alpha} .
\end{array}
$$

Therefore, $B E: B F: B G=\frac{1}{9}: \frac{3}{11}: \frac{1}{3}=11: 27: 33$.
Hence, $B E: E F: F G=11: 16: 6$.