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(A) The position vectors of the vertices $A, B$,and $C$ of the triangle are $a, b$,and $c$ respectively.The position vector of the centroid $G$ of the

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(A) The position vectors of the vertices $A, B$,and $C$ of the triangle are $a, b$,and $c$ respectively.The position vector of the centroid $G$ of the triangle is given by $g = \frac{a + b + c}{3}$.We need to find the sum of the vectors $\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC}$.Using the definition of a vector between two points,$\overrightarrow{GA} = a - g$,$\overrightarrow{GB} = b - g$,and $\overrightarrow{GC} = c - g$.Summing these,we get:$\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC} = (a - g) + (b - g) + (c - g)$$= (a + b + c) - 3g$Substituting $g = \frac{a + b + c}{3}$:$= (a + b + c) - 3 \left( \frac{a + b + c}{3} \right)$$= (a + b + c) - (a + b + c) = 0$.

If $A, B, C$ are the vertices of a triangle whose position vectors are $a, b, c$ and $G$ is the centroid of the $\Delta ABC$,then $\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC}$ is

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