"If the points $\mathrm{P}$ and $\mathrm{Q}$ are respectively the circumcenter and the orthocentre of a $\triangle \mathrm{ABC}$, then $$\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}$$ is equal to :"

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"1. **Circumcenter $ P $**:

The circumcenter of a triangle is equidistant from the vertices of the triangle. It is the center of the circumcircle, the circle that passes through all three vertices of the triangle.

2. **Orthocenter $ Q $**:

The orthocenter of a triangle is the point of intersection of its three altitudes. An altitude of a triangle is a perpendicular line segment drawn from a vertex to its opposite side (or its extension).

3. **Centroid $ G $**:

The centroid of a triangle is the point of intersection of its medians. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. The centroid always divides each median in a 2:1 ratio, with the larger segment being closer to the vertex.

With the above definitions, it's known that the centroid divides the line segment joining the circumcenter and orthocenter in the ratio 2 : 1, meaning :

$ \overrightarrow{PG} = \frac{2}{3} \overrightarrow{PQ} $

$ \overrightarrow{PQ} = 3\overrightarrow{PG} $

The position vector of the centroid $G$ in terms of the vertices $A, B,$ and $C$ is :

$ \overrightarrow{G} = \frac{\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{3} $

Because $P$ is the circumcenter and is at the origin in this problem:

$ \overrightarrow{PA} = \overrightarrow{A} $

$ \overrightarrow{PB} = \overrightarrow{B} $

$ \overrightarrow{PC} = \overrightarrow{C} $

Substituting these into the equation for $G$ :

$ \overrightarrow{PG} = \frac{\overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC}}{3} $

Now, using the relationship between $PQ$ and $PG$ established earlier :

$ \overrightarrow{PQ} = 3\overrightarrow{PG} = \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} $

"

If the points $\mathrm{P}$ and $\mathrm{Q}$ are respectively the circumcenter and the orthocentre of a $\triangle \mathrm{ABC}$, then $$\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}$$ is equal to :

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If the points $\mathrm{P}$ and $\mathrm{Q}$ are respectively the circumcenter and the orthocentre of a $\triangle \mathrm{ABC}$, then $$\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}$$ is equal to :

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