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(A) Let the vertices be $A, B, C$. The centroid $G$ of the triangle is given by the average of the position vectors of its vertices:$G = \frac{(4\over

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$3$

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(A) Let the vertices be $A, B, C$. The centroid $G$ of the triangle is given by the average of the position vectors of its vertices:$G = \frac{(4\overrightarrow{p} + \overrightarrow{q} - 3\overrightarrow{r}) + (-5\overrightarrow{p} + \overrightarrow{q} + 2\overrightarrow{r}) + (2\overrightarrow{p} - \overrightarrow{q} + 2\overrightarrow{r})}{3}$$G = \frac{(4-5+2)\overrightarrow{p} + (1+1-1)\overrightarrow{q} + (-3+2+2)\overrightarrow{r}}{3} = \frac{\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}}{3}$We know that the orthocenter $(O)$,centroid $(G)$,and circumcenter $(C)$ are collinear,and the centroid divides the line segment joining the orthocenter and circumcenter in the ratio $2:1$. Thus,$G = \frac{1 \cdot O + 2 \cdot C}{3}$,which implies $3G = O + 2C$.Given $O = \frac{\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}}{4}$ and $C = \alpha \overrightarrow{p} + \beta \overrightarrow{q} + \gamma \overrightarrow{r}$,we have:$3 \left( \frac{\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}}{3} \right) = \frac{\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}}{4} + 2(\alpha \overrightarrow{p} + \beta \overrightarrow{q} + \gamma \overrightarrow{r})$$\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r} - \frac{1}{4}(\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}) = 2(\alpha \overrightarrow{p} + \beta \overrightarrow{q} + \gamma \overrightarrow{r})$$\frac{3}{4}(\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}) = 2(\alpha \overrightarrow{p} + \beta \overrightarrow{q} + \gamma \overrightarrow{r})$$\alpha \overrightarrow{p} + \beta \overrightarrow{q} + \gamma \overrightarrow{r} = \frac{3}{8}\overrightarrow{p} + \frac{3}{8}\overrightarrow{q} + \frac{3}{8}\overrightarrow{r}$Comparing coefficients,we get $\alpha = \frac{3}{8}, \beta = \frac{3}{8}, \gamma = \frac{3}{8}$.Therefore,$\alpha + 2\beta + 5\gamma = \frac{3}{8} + 2(\frac{3}{8}) + 5(\frac{3}{8}) = \frac{3 + 6 + 15}{8} = \frac{24}{8} = 3$.

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