If the points P and Q are respectively the circumcenter and the orthocentre of a ∆ A B C , then P A…

Let $P$ be origin.

Now from the above triangle we can write $\overrightarrow{PA}=\overrightarrow{a}, \overrightarrow{PA}=\overrightarrow{b}, \overrightarrow{PC}=\overrightarrow{c}$.

Also, we know that centroid $G$ divides orthocentre $Q$ and Circumcentre $P$ in the ratio $2:1$.

Now $\overrightarrow{PG}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}$

$\Rightarrow \overrightarrow{PQ}=3\left(\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}\right)$

$\Rightarrow \overrightarrow{PQ}=\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}$

Therefore, the required answer is $\overrightarrow{PQ}$.