JEE Main 2025 — Mathematics Vectors & 3D Geometry

By given data

$$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{CB}}$$AB+AC=CB

Let pv of $\overrightarrow{\mathrm{A}}$A are $\overrightarrow{\mathrm{O}}$O then

$$\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}$$AB=B−A

i.e. $p v$pv of $\overrightarrow{\mathrm{B}}-=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$B−=2i^−j^​+k^

$$\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{C}}$$CA=A−C

i.e. pv of $\overrightarrow{\mathrm{C}}=-(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$C=−(i^−3j^​−5k^)

Now pv of centroid

$\begin{aligned}
& (\overrightarrow{\mathrm{G}})-=\frac{\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}}{3}=\frac{\overrightarrow{0}+(2,-1,1)+(-1,3,5)}{3} \
& \overrightarrow{\mathrm{G}}=\frac{1}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})
\end{aligned}$

Now $\overrightarrow{\mathrm{AG}}=\frac{1}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})$AG=31​(i^+2j^​+6k^)

$$\Rightarrow|\overrightarrow{\mathrm{AG}}|^2=\frac{1}{9} \times 41$$⇒∣AG∣2=91​×41

$$\overrightarrow{\mathrm{BG}}=\left(\frac{1}{3}-2\right) \hat{\mathrm{i}}+\left(\frac{2}{3}+1\right) \hat{\mathrm{j}}+(2-1) \hat{\mathrm{k}}$$BG=(31​−2)i^+(32​+1)j^​+(2−1)k^

$$\Rightarrow|\overrightarrow{\mathrm{BG}}|^2=\frac{59}{9}$$⇒∣BG∣2=959​

$\begin{aligned}
& \overrightarrow{\mathrm{CG}}=\left(\frac{1}{3}+1\right) \hat{\mathrm{i}}+\left(\frac{2}{3}-3\right) \hat{\mathrm{j}}+(2-5) \hat{\mathrm{k}} \
& \Rightarrow|\overrightarrow{\mathrm{CG}}|^2=\frac{146}{9}
\end{aligned}$

Now

$\begin{aligned}
& 6\left[|\overrightarrow{\mathrm{AG}}|^2+|\overrightarrow{\mathrm{BG}}|^2+|\overrightarrow{\mathrm{CG}}|^2\right]=6 \times\left[\frac{41}{9}+\frac{59}{9}+\frac{146}{9}\right] \
& =6 \times \frac{246}{9}=164
\end{aligned}$