If P, Q , R are the mid-points of the sides AB, BC and CA of `Delta ABC` and O is point whithin the triangle, then `vec (OA) + vec(OB) + vec( OC) =`

Options

A. `2(vec(OP)+vec(OQ) + vec(OR))`
B. `vec(OP)+vec(OQ) + vec(OR)`
C. `4(vec(OP)+vec(OQ) + vec(OR))`
D. `6(vec(OP)+vec(OQ) + vec(OR))`

Answer

To solve the problem, we need to find the expression for \( \vec{OA} + \vec{OB} + \vec{OC} \) where \( O \) is a point within triangle \( ABC \) and \( P, Q, R \) are the midpoints of sides \( AB, BC, \) and \( CA \) respectively.

### Step-by-Step Solution:

1. **Define the Position Vectors**:
Let the position vectors of points \( A, B, C \) be represented as \( \vec{A}, \vec{B}, \vec{C} \). The position vector of point \( O \) is \( \vec{O} \).

2. **Express the Vectors**:
The vectors from point \( O \) to points \( A, B, C \) can be expressed as:
\[ \vec{OA} = \vec{A} - \vec{O}, \quad \vec{OB} = \vec{B} - \vec{O}, \quad \vec{OC} = \vec{C} - \vec{O} \]

3. **Sum the Vectors**:
Now, we can sum these vectors:
\[ \vec{OA} + \vec{OB} + \vec{OC} = (\vec{A} - \vec{O}) + (\vec{B} - \vec{O}) + (\vec{C} - \vec{O}) \]

4. **Combine Like Terms**:
Combine the terms:
\[ \vec{OA} + \vec{OB} + \vec{OC} = \vec{A} + \vec{B} + \vec{C} - 3\vec{O} \]

5. **Final Expression**:
Thus, we can express the result as:
\[ \vec{OA} + \vec{OB} + \vec{OC} = \vec{A} + \vec{B} + \vec{C} - 3\vec{O} \]

### Conclusion:
The final result for \( \vec{OA} + \vec{OB} + \vec{OC} \) is:
\[ \vec{OA} + \vec{OB} + \vec{OC} = \vec{A} + \vec{B} + \vec{C} - 3\vec{O} \]