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Q.
Let the position vectors of the points $A, B$ and $C$ be $a , b$ and $c$, respectively. Let $Q$ be the point of intersection of the medians of the $\Delta A B C .$ Then, $Q A + Q B + Q C$ is equal to
Since, $O$ is the intersecting point of all the medians of $\triangle A B C$.
Hence, $ Q =\frac{ a + b + c }{3}$
Now, $d Q A = a -\frac{ a + b + c }{3}$
$ =\frac{2 a - b - c }{3} $
{ Similarly, $ Q B =\frac{2 b - a - c }{3} $
and $ Q C =\frac{2 c - a - b }{3}$
Hence, $Q A + Q B + Q C =\frac{2 a - b - c }{3} $
$+\frac{2 b - a - c }{3}+\frac{2 c - a - b }{3}=0 $
