The vertices of a triangle have the position vectors `veca,vecb,vecc` and `p( r)` is a point in the plane of `Delta` such that: `veca.vecb+ vecc.vecr = veca.vecc + vecb.vecr = vecb.vecc+veca.vecr` then for the `Delta`, P is the:

Options

A. circumcentre
B. centroid
C. orthocentre
D. incentre

Answer

To solve the problem, we need to analyze the given equations involving the position vectors of the vertices of the triangle and the point \( P \) in the plane of the triangle.

### Step-by-Step Solution:

1. **Given Equations**:
We have the following equations:
\[ \vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{r} = \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{r} = \vec{b} \cdot \vec{c} + \vec{a} \cdot \vec{r} \]
Let's denote the common value of these expressions as \( k \).

2. **First Equation**:
Start with the first part of the equation:
\[ \vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{r} = k \]
Rearranging gives:
\[ \vec{c} \cdot \vec{r} = k - \vec{a} \cdot \vec{b} \]

3. **Second Equation**:
Now consider the second part:
\[ \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{r} = k \]
Rearranging gives:
\[ \vec{b} \cdot \vec{r} = k - \vec{a} \cdot \vec{c} \]

4. **Third Equation**:
Now look at the third part:
\[ \vec{b} \cdot \vec{c} + \vec{a} \cdot \vec{r} = k \]
Rearranging gives:
\[ \vec{a} \cdot \vec{r} = k - \vec{b} \cdot \vec{c} \]

5. **Setting Up Perpendicularity**:
From the rearranged equations, we can express the relationships between the vectors. We can analyze these equations to find conditions for perpendicularity:
- From the first equation, we can derive that \( \vec{a} \) is perpendicular to \( \vec{b} - \vec{c} \).
- From the second equation, we can derive that \( \vec{c} \) is perpendicular to \( \vec{a} - \vec{b} \).
- From the third equation, we can derive that \( \vec{b} \) is perpendicular to \( \vec{a} - \vec{c} \).

6. **Conclusion**:
Since all three conditions indicate that the point \( P \) is the intersection of the altitudes of the triangle, we conclude that point \( P \) is the orthocenter of triangle \( \Delta ABC \).

### Final Answer:
Thus, the point \( P \) is the **orthocenter** of triangle \( \Delta ABC \).