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 $\dfrac{\vec p+\vec q+\vec r}{4}$ and $\alpha \vec p+\beta \vec q+\gamma \vec r$ respectively, then $\alpha+2\beta+5\gamma$ is equal to: