9. Given $0<\lambda<1$. Planar vectors $\boldsymbol{a} 、 \boldsymbol{b} 、 \boldsymbol{c}$ satisfy
$$
|a|=1,|b|=2,|c|=3 \text {. }
$$

If $\boldsymbol{b} \cdot \boldsymbol{c}=0$, then the set of all values that $|\boldsymbol{a}-\lambda \boldsymbol{b}-(1-\lambda) \boldsymbol{c}|$ cannot take is $\qquad$ .
9. $\left[\frac{6 \sqrt{13}}{13}-1,4\right]$.

Translate the vectors $\boldsymbol{b}$ and $\boldsymbol{c}$ to the origin $O$, and establish a Cartesian coordinate system with $b$ and $c$ as the positive directions of the $x$-axis and $y$-axis, respectively. Then, the coordinates of the point corresponding to the vector $\lambda \boldsymbol{b}+(1-\lambda) \boldsymbol{c}$ are $P(2 \lambda, 3(1-\lambda))$.
Thus, $O P=\sqrt{13 \lambda^{2}-18 \lambda+9}$.
And $|\boldsymbol{a}-\lambda \boldsymbol{b}-(1-\lambda) \boldsymbol{c}|$ represents the distance $d$ from point $P$ to the unit circle, so the upper bound of $d$ is 4, and the lower bound is $\frac{6 \sqrt{13}}{13}-1$