JEE Main 2025 — 4 April, Evening Shift — Question 41

Assuming Vertex $A$ to be origin

$\vec{A}=\vec{a}_{1}=\overrightarrow{0}$

$\vec{B}=\vec{a}_{1}+\vec{u}=\vec{u}=2 \hat{i}-\hat{j}+\hat{k}$

$\vec{C}=\vec{a}_{1}+\vec{v}=\vec{v}=3 \hat{i}-4 \hat{j}-4 \hat{k}$

One solving

$\vec{A}=\overrightarrow{0}, \vec{B}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{C}=3 \hat{i}-4 \hat{j}-4 \hat{k}$, are the position vector of vertices $A B$ and $C$ respectively.

$\vec{G}=\frac{1}{3}(\vec{A}+\vec{B}+\vec{C})=\frac{1}{3}(\overrightarrow{0}+\vec{B}+\vec{C})=\frac{1}{3}(\vec{B}+\vec{C})$

$\Rightarrow \quad \vec{G}=\frac{5}{3} \hat{i}-\frac{5}{3} \hat{j}-\hat{k}$

$\overrightarrow{A G}=\vec{G}-\vec{A}=\vec{G}$

$|\overrightarrow{A G}|^{2}=\left(\frac{5}{3}\right)^{2}+\left(\frac{5}{3}\right)^{2}+(1)^{2}=\frac{25}{9}+\frac{25}{9}+1=\frac{50}{9}+1=\frac{59}{9}$

$\overrightarrow{B C}=\vec{G}-\vec{B}$

$\vec{B}=2 \hat{i}-\hat{j}+\hat{k}$

$|\overrightarrow{B G}|^{2}=\left(\frac{1}{3}\right)^{3}+\left(\frac{2}{3}\right)^{2}+4=\frac{1}{9}+\frac{4}{9}+4=\frac{5}{9}+4=\frac{41}{9}$

$\overrightarrow{C G}=\vec{G}-\vec{C}$ $\vec{C}=3 \hat{i}-4 \hat{j}-4 \hat{k}$

$\left.\overrightarrow{C G}\right|^{2}=\left(\frac{4}{3}\right)^{2}+\left(\frac{7}{3}\right)^{2}+9=\frac{16}{9}+\frac{49}{9}+9=\frac{65}{9}+9=\frac{65}{9}+\frac{81}{9}=\frac{146}{9}$

$6\left(|\overrightarrow{A G}|^{2}+|\overrightarrow{B G}|^{2}+|\overrightarrow{C G}|^{2}\right)=6 \cdot\left(\frac{59}{9}+\frac{41}{9}+\frac{146}{9}\right)=6 \cdot \frac{246}{9}=164$