6. In $\triangle A B C$, $\angle B A C=120^{\circ}, A B=2$, $A C=1, D$ is a point on side $B C$ (including endpoints). Then the range of $\overrightarrow{A D} \cdot \overrightarrow{B C}$ is ( ).
(A) $[2,5]$
(B) $[-5,3]$
(C) $[0,5]$
(D) $[-5,2]$
6. D.

Let $\overrightarrow{B D}=x \overrightarrow{B C}(0 \leqslant x \leqslant 1)$. Then $\overrightarrow{A D}=(1-x) \overrightarrow{A B}+x \overrightarrow{A C}$. Therefore, $\overrightarrow{A D} \cdot \overrightarrow{B C}=7 x-5 \in[-5,2]$.