Let `P` be a point interior to the acute triangle `A B Cdot` If `P A+P B+P C` is a null vector, then w.r.t traingel `A B C ,` point `P` is its a. centroid b. orthocentre c. incentre d. circumcentre

Options

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

Explanation

To solve the problem, we need to analyze the given condition that the sum of the vectors from point \( P \) to the vertices \( A \), \( B \), and \( C \) of triangle \( ABC \) is a null vector. Let's break this down step by step.

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We are given that \( \vec{PA} + \vec{PB} + \vec{PC} = \vec{0} \). This means that the vector sum of the position vectors from point \( P \) to each vertex of triangle \( ABC \) results in the null vector.

2. **Expressing the Vectors**:
Let the position vectors of points \( A \), \( B \), \( C \), and \( P \) be represented as \( \vec{a} \), \( \vec{b} \), \( \vec{c} \), and \( \vec{p} \) respectively. The vectors \( \vec{PA} \), \( \vec{PB} \), and \( \vec{PC} \) can be expressed as:
\[ \vec{PA} = \vec{a} - \vec{p}, \quad \vec{PB} = \vec{b} - \vec{p}, \quad \vec{PC} = \vec{c} - \vec{p} \]

3. **Substituting the Vectors into the Equation**:
Substitute the expressions for \( \vec{PA} \), \( \vec{PB} \), and \( \vec{PC} \) into the given equation:
\[ (\vec{a} - \vec{p}) + (\vec{b} - \vec{p}) + (\vec{c} - \vec{p}) = \vec{0} \]

4. **Simplifying the Equation**:
Combine the terms:
\[ \vec{a} + \vec{b} + \vec{c} - 3\vec{p} = \vec{0} \]
Rearranging gives us:
\[ \vec{a} + \vec{b} + \vec{c} = 3\vec{p} \]

5. **Finding the Position Vector of Point \( P \)**:
To find \( \vec{p} \), divide both sides by 3:
\[ \vec{p} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} \]

6. **Identifying the Geometric Interpretation**:
The expression \( \vec{p} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} \) represents the centroid of triangle \( ABC \).

### Conclusion:
Thus, point \( P \) is the centroid of triangle \( ABC \).

### Answer:
**a. centroid**