If the points P and Q are respectively the circumcenter and the...

Let $P$ be origin.

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

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

Now $\overset{\to}{PG} = \dfrac{\overset{\to}{a} + \overset{\to}{b} + \overset{\to}{c}}{3}$

$⇒ \overset{\to}{PQ} = 3\left(\dfrac{\overset{\to}{a} + \overset{\to}{b} + \overset{\to}{c}}{3}\right)$

$⇒ \overset{\to}{PQ} = \overset{\to}{a} + \overset{\to}{b} + \overset{\to}{c} = \overset{\to}{PA} + \overset{\to}{PB} + \overset{\to}{PC}$

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