Answer
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Hint: In triangle ABC we can define centroid as $\overrightarrow{G}=\dfrac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3}$ where $\overrightarrow{A},\overrightarrow{B},\overrightarrow{C}$ are sides vector of triangle ABC. If have length of sides of triangle then we can use it as in same form.
Complete step by step solution:
If sides vectors of triangle ABC are $\overrightarrow{A},\overrightarrow{B},\overrightarrow{C}$. Then we can write centroid of triangle as
$\overrightarrow{G}=\dfrac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3}$
We can arrange it as
$3\overrightarrow{G}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$..................................(i)
We can also write
$\overrightarrow{GA}=\overrightarrow{A}-\overrightarrow{G}$
$\overrightarrow{GB}=\overrightarrow{B}-\overrightarrow{G}$
$\overrightarrow{GC}=\overrightarrow{C}-\overrightarrow{G}$
Hence given expression can be written as
\[\Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{A}-\overrightarrow{G}+\overrightarrow{B}-\overrightarrow{G}+\overrightarrow{C}-\overrightarrow{G}\]
\[\Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}-3\overrightarrow{G}\]
From equation (i) we can write value of \[\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}\]
\[\Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=3\overrightarrow{G}-3\overrightarrow{G}\]
\[\Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0\]
Hence option a is correct.
Note: We can use the same formula of centroid in terms of coordinate as well. Also we can write $\overrightarrow{AB}$ as difference of individual vector A and vector B as below:
$\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}$
But we always write vector B first and then vector A. We need to remember this point.
Complete step by step solution:
If sides vectors of triangle ABC are $\overrightarrow{A},\overrightarrow{B},\overrightarrow{C}$. Then we can write centroid of triangle as
$\overrightarrow{G}=\dfrac{\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}}{3}$
We can arrange it as
$3\overrightarrow{G}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$..................................(i)
We can also write
$\overrightarrow{GA}=\overrightarrow{A}-\overrightarrow{G}$
$\overrightarrow{GB}=\overrightarrow{B}-\overrightarrow{G}$
$\overrightarrow{GC}=\overrightarrow{C}-\overrightarrow{G}$
Hence given expression can be written as
\[\Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{A}-\overrightarrow{G}+\overrightarrow{B}-\overrightarrow{G}+\overrightarrow{C}-\overrightarrow{G}\]
\[\Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}-3\overrightarrow{G}\]
From equation (i) we can write value of \[\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}\]
\[\Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=3\overrightarrow{G}-3\overrightarrow{G}\]
\[\Rightarrow \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0\]
Hence option a is correct.
Note: We can use the same formula of centroid in terms of coordinate as well. Also we can write $\overrightarrow{AB}$ as difference of individual vector A and vector B as below:
$\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}$
But we always write vector B first and then vector A. We need to remember this point.
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