JEE Main 2025 — 24 January, Evening Shift — Question 6

We know that O (orthocentre) $\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4}$

$C$ (circum centre) $\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}}$

$C($ centroid $)=\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3}$

by relation

$\Rightarrow 2(\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}})+\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4}=3\left(\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3}\right)$

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

$\Rightarrow 8 \alpha=3,8 \beta=3,8 \gamma=3$

$\alpha=\frac{3}{8}, \beta=\frac{3}{8}, \gamma=\frac{3}{8}$

$\therefore \alpha+2 \beta+5 \gamma$

$\frac{3}{8}+\frac{6}{8}+\frac{15}{8}=\frac{24}{8}=3$

$$\boxed3$$