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(C) Let $\vec{a'}, \vec{b'}, \vec{c'}$ be the position vectors of the vertices $A', B', C'$ respectively.Since lines are drawn through the vertices pa

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$\frac{\vec{a}+\vec{b}+\vec{c}}{3}$

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(C) Let $\vec{a'}, \vec{b'}, \vec{c'}$ be the position vectors of the vertices $A', B', C'$ respectively.Since lines are drawn through the vertices parallel to the opposite sides,$A$ is the midpoint of $B'C'$,$B$ is the midpoint of $A'C'$,and $C$ is the midpoint of $A'B'$.Using the midpoint formula:$\vec{a} = \frac{\vec{b'} + \vec{c'}}{2} \implies \vec{b'} + \vec{c'} = 2\vec{a}$$\vec{b} = \frac{\vec{a'} + \vec{c'}}{2} \implies \vec{a'} + \vec{c'} = 2\vec{b}$$\vec{c} = \frac{\vec{a'} + \vec{b'}}{2} \implies \vec{a'} + \vec{b'} = 2\vec{c}$Adding these three equations:$2(\vec{a'} + \vec{b'} + \vec{c'}) = 2(\vec{a} + \vec{b} + \vec{c})$$\vec{a'} + \vec{b'} + \vec{c'} = \vec{a} + \vec{b} + \vec{c}$The centroid of $\Delta A'B'C'$ is given by $G' = \frac{\vec{a'} + \vec{b'} + \vec{c'}}{3} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$.

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices of a triangle $ABC$. Through the vertices,lines are drawn parallel to the sides to form the triangle $A'B'C'$. Then the centroid of $\Delta A'B'C'$ is

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