We know that,

$$\begin{aligned} &\begin{aligned} & \mathrm{O} \text { (orthocentre) } \frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4} \\ & \mathrm{C}(\text { circum centre) } \alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}} \\ & \mathrm{C} \text { (centroid) }=\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3} \end{aligned}\\ &\text { by relation }\\ &\begin{aligned} & \Rightarrow 2(\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}})+\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{4}=3\left(\frac{\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}}{3}\right) \\ & \Rightarrow 8(\alpha \overrightarrow{\mathrm{p}}+\beta \overrightarrow{\mathrm{q}}+\gamma \overrightarrow{\mathrm{r}})=3(\overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}}+\overrightarrow{\mathrm{r}}) \\ & \Rightarrow 8 \alpha=3,8 \beta=3,8 \gamma=3 \\ & \alpha=\frac{3}{8}, \beta=\frac{3}{8}, \gamma=\frac{3}{8} \\ & \therefore \alpha+2 \beta+3 \gamma \\ & \frac{3}{8}+\frac{6}{8}+\frac{15}{8}=\frac{24}{8}=3 \end{aligned} \end{aligned}$$