9. Given planar vectors $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ satisfying $|\boldsymbol{a}|=1,|\boldsymbol{b}|=2,|\boldsymbol{c}|=3,0<\lambda<1$. If $\boldsymbol{b} \cdot \boldsymbol{c}=0$, then the values that $|\boldsymbol{a}-\lambda \boldsymbol{b}-(1-\lambda) \boldsymbol{c}|$ cannot take are $\qquad$ .
As shown in the figure, $P$ is a point on line segment $B C$, then $\overrightarrow{O P}=\lambda \boldsymbol{b}+(1-\lambda) \boldsymbol{c}$, $0<\lambda<1$, hence $\overrightarrow{P A}=\boldsymbol{a}-\lambda \boldsymbol{b}-(1-\lambda) \boldsymbol{c}$. Since the minimum distance from point $P$ on line segment $B C$ to point $A$ on circle $O$ is $\frac{6}{\sqrt{13}}-1$, and the maximum distance is $3+1=4$, the values that $|\boldsymbol{a}-\lambda \boldsymbol{b}-(1-\lambda) \boldsymbol{c}|$ cannot take are $\left(-\infty, \frac{6}{\sqrt{13}}-1\right) \cup(4,+\infty)$