24. Given that quadrilateral $ABCD$ is inscribed in $\odot O$, and $AB = 3, BC = 2, CD = 6, DA = 8$, points $X$ and $Y$ are on side $BD$, and $\frac{DX}{BD} = \frac{1}{4}, \frac{BY}{BD} = \frac{11}{36}$. A line through point $Y$ parallel to $AD$ intersects $AX$ at point $E$. A line through point $E$ parallel to $AC$ intersects $CX$ at point $F$. Let the intersection of line $CX$ with $\odot O$ other than $C$ be $G$. Then the value of $XF \cdot XG$ is ( ).
(A) 17
(B) $\frac{59-5 \sqrt{2}}{3}$
(C) $\frac{91-12 \sqrt{3}}{4}$
(D) 18
(E) $\frac{67-10 \sqrt{2}}{3}$
Obviously, $\frac{X Y}{B D}=1-\frac{1}{4}-\frac{11}{36}=\frac{4}{9}$.
Thus, $\frac{D X}{X Y}=\frac{9}{16}$.
Given $A D / / E Y, E F / / A C$
$$
\Rightarrow \frac{A X}{X E}=\frac{D X}{X Y}=\frac{X C}{F X}=\frac{9}{16} \text {. }
$$

Now, let's find the length of $B D$.
From the fact that points $A, B, C, D$ are concyclic and using the cosine rule, we have
$$
\begin{array}{l}
B D^{2}=3^{2}+8^{2}-48 \cos \angle B A D \\
=2^{2}+6^{2}-24 \cos (\pi-\angle B A D) \\
=2^{2}+6^{2}+24 \cos \angle B A D \\
\Rightarrow \cos \angle B A D=\frac{11}{24}, B D=\sqrt{51} .
\end{array}
$$

Also, $C X \cdot X G=D X \cdot X B=\frac{\sqrt{51}}{4} \times \frac{3 \sqrt{51}}{4}$, so
$$
X F \cdot X G=\frac{X F}{C X} C X \cdot X G=\frac{51}{3}=17
$$