Let the position vectors of three vertices of a triangle be $$4\vec p+\vec q-3\vec r,\;-5\vec p+\vec q+2\vec r$$ and $$2\vec p-\vec q+2\vec r.$$ If the position vectors of the orthocenter and the circumcenter  of the triangle are $$\f\frac{\vec p+\vec q+\vec r}{4}$$ and $$\alpha\vec p+\beta\vec q+\gamma\vec r$$ { respectively, then $$\alpha+2\beta+5\gamma$$  is equal to:

Let the vertices of the triangle be $$P = 4\vec{p} + \vec{q} - 3\vec{r}$$, $$Q = -5\vec{p} + \vec{q} + 2\vec{r}$$, $$R = 2\vec{p} - \vec{q} + 2\vec{r}$$.

The orthocenter is $$H = \f\frac{\vec{p} + \vec{q} + \vec{r}}{4}$$.

We use the property that the centroid $$G$$ divides $$OH$$ in ratio $$1:2$$ (where $$O$$ is circumcenter), i.e., $$G = \f\frac{O + 2 \cdot \text{(midpoint)}...}$$. More precisely: $$\vec{G} = \f\frac{\vec{P}+\vec{Q}+\vec{R}}{3}$$.

$$\vec{G} = \f\frac{(4-5+2)\vec{p} + (1+1-1)\vec{q} + (-3+2+2)\vec{r}}{3} = \f\frac{\vec{p} + \vec{q} + \vec{r}}{3}$$

Euler line relation: $$\vec{G} = \f\frac{\vec{H} + 2\vec{O}}{3}$$ where $$O$$ is the circumcenter.

$$\f\frac{\vec{p}+\vec{q}+\vec{r}}{3} = \f\frac{1}{3}\l\left(\f\frac{\vec{p}+\vec{q}+\vec{r}}{4} + 2\vec{O}\r\right)$$

$$\vec{p}+\vec{q}+\vec{r} = \f\frac{\vec{p}+\vec{q}+\vec{r}}{4} + 2\vec{O}$$

$$2\vec{O} = \f\frac{3(\vec{p}+\vec{q}+\vec{r})}{4}$$

$$\vec{O} = \f\frac{3}{8}(\vec{p}+\vec{q}+\vec{r}) = \f\frac{3}{8}\vec{p} + \f\frac{3}{8}\vec{q} + \f\frac{3}{8}\vec{r}$$

So $$\alpha = 3/8$$, $$\beta = 3/8$$, $$\gamma = 3/8$$.

$$\alpha + 2\beta + 5\gamma = \f\frac{3}{8} + \f\frac{6}{8} + \f\frac{15}{8} = \f\frac{24}{8} = 3$$.

The correct answer is Option A: 3.