6. Given three non-collinear unit vectors $\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}$ on a plane, satisfying $\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c}=\boldsymbol{0}$. If $0 \leqslant t \leqslant 1$, then the range of $|-2 a+t b+(1-t) c|$ is $\qquad$
6. $\left[\frac{5}{2}, \sqrt{7}\right]$.

Notice, $-2 a=2 b+2 c$.
$$
\begin{array}{c}
\text { Then }-2 \boldsymbol{a}+t \boldsymbol{b}+(1-t) \boldsymbol{c} \\
=(t+2) \boldsymbol{b}+(3-t) \boldsymbol{c} \text {. }
\end{array}
$$

Therefore, $|-2 a+t b+(1-t) c|$
$$
\begin{array}{l}
=\sqrt{(t+2)^{2}+(3-t)^{2}+2(t+2)(3-t) \boldsymbol{b} \cdot \boldsymbol{c}} \\
=\sqrt{2 t^{2}-2 t+13+2\left(-t^{2}+t+6\right)\left(-\frac{1}{2}\right)} \\
=\sqrt{3 t^{2}-3 t+7} \in\left[\frac{5}{2}, \sqrt{7}\right] .
\end{array}
$$