JEE Main 2025 — Vector Algebra Question with Solution

By given data
$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{CB}}$
Let pv of $\overrightarrow{\mathrm{A}}$ are $\overrightarrow{\mathrm{O}}$ then
$\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}$
i.e. $pv$ of $\overrightarrow{\mathrm{B}}-=2\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{C}}$
i.e. pv of $\overrightarrow{\mathrm{C}}=-(\hat{\mathrm{i}}-3\hat{\mathrm{j}}-5\hat{\mathrm{k}})$
Now pv of centroid
$\begin{aligned}
& (\overrightarrow{\mathrm{G}})-=\frac{\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}}{3}=\frac{\overrightarrow{0}+(2,-1,1)+(-1,3,5)}{3} \\ & \overrightarrow{\mathrm{G}}=\frac{1}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})
\end{aligned}$<br/>Now$\overrightarrow{\mathrm{AG}}=\frac{1}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})$<br/>$\Rightarrow|\overrightarrow{\mathrm{AG}}|^2=\frac{1}{9} \times 41$<br/>$\overrightarrow{\mathrm{BG}}=\left(\frac{1}{3}-2\right) \hat{\mathrm{i}}+\left(\frac{2}{3}+1\right) \hat{\mathrm{j}}+(2-1) \hat{\mathrm{k}}$<br/>$\Rightarrow|\overrightarrow{\mathrm{BG}}|^2=\frac{59}{9}$<br/>$\begin{aligned}
& \overrightarrow{\mathrm{CG}}=\left(\frac{1}{3}+1\right) \hat{\mathrm{i}}+\left(\frac{2}{3}-3\right) \hat{\mathrm{j}}+(2-5) \hat{\mathrm{k}} \\ & \Rightarrow|\overrightarrow{\mathrm{CG}}|^2=\frac{146}{9}
\end{aligned}$<br/>Now<br/>$\begin{aligned}
& 6\left[|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right]=6 \times\left[\frac{41}{9}+\frac{59}{9}+\frac{146}{9}\right] \\ & =6 \times \frac{246}{9}=164
\end{aligned}$