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 $\frac{\vec{p} + \vec{q} + \vec{r}}{2}$ and $\alpha \vec{p} + \beta \vec{q} + \gamma \vec{r}$ respectively, then $\alpha + 2 \beta + 5 \gamma$ is equal to:

Step 1: Define position vectors of vertices, orthocenter, and circumcenter

We are given the position vectors of the three vertices of the triangle, denoted as $\vec{A}$, $\vec{B}$, and $\vec{C}$. We are also given the position vector of the orthocenter, $\vec{H}$, and the circumcenter, $\vec{O}$. We need to find the values of $\alpha$, $\beta$, and $\gamma$ from the circumcenter's position vector.

Step 2: Apply Euler's Line Theorem

For any triangle, the orthocenter (H), centroid (G), and circumcenter (O) are collinear, and the centroid divides the segment HO in the ratio 2:1. This relationship is known as Euler's Line Theorem, which states that $\vec{H} + 2\vec{O} = 3\vec{G}$. Since the centroid $\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}$, we can substitute this into the Euler's Line equation to get $\vec{H} + 2\vec{O} = \vec{A} + \vec{B} + \vec{C}$.

Step 3: Substitute known vectors into Euler's Line equation

Now, we substitute the given position vectors for $\vec{A}$, $\vec{B}$, $\vec{C}$, $\vec{H}$, and $\vec{O}$ into the Euler's Line equation. This will allow us to form an equation in terms of $\vec{p}$, $\vec{q}$, and $\vec{r}$.

Step 4: Simplify and equate coefficients

We simplify both sides of the equation by collecting the coefficients of $\vec{p}$, $\vec{q}$, and $\vec{r}$. On the right side, $(4-5+2)\vec{p} = \vec{p}$, $(1+1-1)\vec{q} = \vec{q}$, and $(-3+2+2)\vec{r} = \vec{r}$. Since $\vec{p}$, $\vec{q}$, and $\vec{r}$ are non-coplanar vectors, their coefficients on both sides of the equation must be equal.

Step 5: Solve for $\alpha$, $\beta$, and $\gamma$

Equating the coefficients of $\vec{p}$, $\vec{q}$, and $\vec{r}$ from the simplified equation, we solve for $\alpha$, $\beta$, and $\gamma$. In this case, all three coefficients turn out to be $\frac{1}{4}$.

Step 6: Calculate $\alpha + 2\beta + 5\gamma$

Finally, we substitute the calculated values of $\alpha$, $\beta$, and $\gamma$ into the expression $\alpha + 2\beta + 5\gamma$ to find the final answer.

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