JEE Main 2025 Jan 24 Shift 2, Mathematics Q6: Applications of Vectors

Two standard facts about a triangle:

• The centroid $G$ divides $OC$ (orthocentre $O$ to circumcentre $C$) in the ratio $OG:GC=2:1$, so $\vec{G}=\frac{2\vec{C}+\vec{O}}{3}$, equivalently $3\vec{G}=2\vec{C}+\vec{O}$.
• The centroid is the average of the three vertices.

Compute the centroid from the given vertices:

$\vec{G}=\frac{(4\vec{p}+\vec{q}-3\vec{r})+(-5\vec{p}+\vec{q}+2\vec{r})+(2\vec{p}-\vec{q}+2\vec{r})}{3}=\frac{\vec{p}+\vec{q}+\vec{r}}{3}$

Substitute into $3\vec{G}=2\vec{C}+\vec{O}$:

$\vec{p}+\vec{q}+\vec{r}=2(\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r})+\frac{\vec{p}+\vec{q}+\vec{r}}{4}$

Multiply through by 4 and collect:

$4(\vec{p}+\vec{q}+\vec{r})=8(\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r})+(\vec{p}+\vec{q}+\vec{r})$

$3(\vec{p}+\vec{q}+\vec{r})=8(\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r})$

Since $\vec{p},\vec{q},\vec{r}$ are independent: $8\alpha =8\beta =8\gamma =3$, so $\alpha =\beta =\gamma =\frac{3}{8}$.

$\alpha +2\beta +5\gamma =\frac{3}{8}(1+2+5)=\frac{3\times 8}{8}=3$

Correct option: (1)