设 $e_{1}$ 与 $e_{2}$ 是两个不共线向量, $\overrightarrow{A B}=3 e_{1}+2 e_{2}, \overrightarrow{C B}=k e_{1}+e_{2}$, $\overrightarrow{C D}=3 e_{1}-2 k e_{2}$, 若 $A, B, D$ 三点共线, 则 $k$ 的值为 \$ \qquad \$
因为 $\overrightarrow{A B}=3 e_{1}+2 e_{2}, \overrightarrow{C B}=k e_{1}+e_{2}, \overrightarrow{C D}=3 e_{1}-2 k e_{2}$,

所以 $\stackrel{\text { ulu }}{B D}=\stackrel{\text { ulut }}{C D}-\stackrel{\text { ulu }}{C D}=3 e_{1}-2 k e_{2}-\left(k e_{1}+e_{2}\right)=(3-k) e_{1}-(2 k+1) e_{2}$,

由题意, $A, B, D$ 三点共线, 故必存在一个实数 $\lambda$, 使得 $\overrightarrow{A B}=\lambda \overrightarrow{B D}$,

所以 $3 e_{1}+2 e_{2}=\lambda(3-k) e_{1}-\lambda(2 k+1) e_{2}$,

又因为 $e_{1}$ 与 $e_{2}$ 不共线, 所以 $\left\{\begin{array}{l}3=\lambda(3-k) \\ 2=-\lambda(2 k+1)\end{array}\right.$, 解得 $k=-\frac{9}{4}$,

故答案为 $-\frac{9}{4}$.