Let the position vectors of three vertices of a triangle be \(4 \vec{p}+\vec{q}-3 \vec{r},-5 \vec{p}+\vec{q}+2 \vec{r}\)…

\(\text{Given, the position vector of triangle are}\\ 4\vec{\mathrm{p}}+\vec{\mathrm{q}}-3\vec{\mathrm{r}},-5\vec{\mathrm{p}}+\vec{\mathrm{q}}+2\vec{\mathrm{r}}\\ &2\vec{\mathrm{p}}-\vec{\mathrm{q}}+2\vec{\mathrm{r}}\\ \mathrm{We}\mathrm{know}\mathrm{that}\\ OG:GC=2:1\\ \mathrm{O}\text{ (orthocentre) =}\frac{\vec{\mathrm{p}}+\vec{\mathrm{q}}+\vec{\mathrm{r}}}{4}\\ \mathrm{C}\text{ (circum centre)= }\alpha \vec{\mathrm{p}}+\beta \vec{\mathrm{q}}+\gamma \vec{\mathrm{r}}\\ G\left(\mathrm{centroid}\right)=\frac{\vec{\mathrm{p}}+\vec{\mathrm{q}}+\vec{\mathrm{r}}}{3}\\ \text{By relation, }\\ \Rightarrow 2\left(\alpha \vec{\mathrm{p}}+\beta \vec{\mathrm{q}}+\gamma \vec{\mathrm{r}}\right)+\frac{\vec{\mathrm{p}}+\vec{\mathrm{q}}+\vec{\mathrm{r}}}{4}=3\left(\frac{\vec{\mathrm{p}}+\vec{\mathrm{q}}+\vec{\mathrm{r}}}{3}\right)\\ \Rightarrow 8\left(\alpha \vec{\mathrm{p}}+\beta \vec{\mathrm{q}}+\gamma \vec{\mathrm{r}}\right)=3\left(\vec{\mathrm{p}}+\vec{\mathrm{q}}+\vec{\mathrm{r}}\right)\\ \text{On comparing the coefficient, we get}\\ \Rightarrow 8\alpha =3,8\beta =3,8\gamma =3\\ \alpha =\frac{3}{8},\beta =\frac{3}{8},\gamma =\frac{3}{8}\\ ∴\alpha +2\beta +5\gamma =\frac{3}{8}+\frac{6}{8}+\frac{15}{8}=\frac{24}{8}=3\)