D, E and F are the mid-points of the sides BC, CA and AB respectively of `Delta ABC` and G is the centroid of the triangle, then `vec(GD) + vec(GE) + vec(GF) = `

Options

A. `vec 0`
B. `2 vec(AB)`
C. `2 vec(GA)`
D. `2 vec(GC)`

Answer

To solve the problem, we need to find the sum of the vectors \( \vec{GD} + \vec{GE} + \vec{GF} \), where \( D, E, F \) are the midpoints of the sides \( BC, CA, \) and \( AB \) respectively of triangle \( ABC \), and \( G \) is the centroid of the triangle.

### Step-by-Step Solution:

1. **Identify the Points**:
Let the position vectors of points \( A, B, C \) be represented as \( \vec{A}, \vec{B}, \vec{C} \).

2. **Find the Midpoints**:
The midpoints \( D, E, F \) can be calculated as follows:
- \( \vec{D} = \frac{\vec{B} + \vec{C}}{2} \)
- \( \vec{E} = \frac{\vec{C} + \vec{A}}{2} \)
- \( \vec{F} = \frac{\vec{A} + \vec{B}}{2} \)

3. **Find the Centroid**:
The centroid \( G \) of triangle \( ABC \) is given by:
\[ \vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} \]

4. **Calculate \( \vec{GD} \)**:
\[ \vec{GD} = \vec{D} - \vec{G} = \left(\frac{\vec{B} + \vec{C}}{2}\right) - \left(\frac{\vec{A} + \vec{B} + \vec{C}}{3}\right) \]
To simplify this, find a common denominator:
\[ \vec{GD} = \frac{3(\vec{B} + \vec{C}) - 2(\vec{A} + \vec{B} + \vec{C})}{6} = \frac{3\vec{B} + 3\vec{C} - 2\vec{A} - 2\vec{B} - 2\vec{C}}{6} = \frac{\vec{B} + \vec{C} - 2\vec{A}}{6} \]

5. **Calculate \( \vec{GE} \)**:
\[ \vec{GE} = \vec{E} - \vec{G} = \left(\frac{\vec{C} + \vec{A}}{2}\right) - \left(\frac{\vec{A} + \vec{B} + \vec{C}}{3}\right) \]
Again, find a common denominator:
\[ \vec{GE} = \frac{3(\vec{C} + \vec{A}) - 2(\vec{A} + \vec{B} + \vec{C})}{6} = \frac{3\vec{C} + 3\vec{A} - 2\vec{A} - 2\vec{B} - 2\vec{C}}{6} = \frac{\vec{A} + \vec{C} - 2\vec{B}}{6} \]

6. **Calculate \( \vec{GF} \)**:
\[ \vec{GF} = \vec{F} - \vec{G} = \left(\frac{\vec{A} + \vec{B}}{2}\right) - \left(\frac{\vec{A} + \vec{B} + \vec{C}}{3}\right) \]
Find a common denominator:
\[ \vec{GF} = \frac{3(\vec{A} + \vec{B}) - 2(\vec{A} + \vec{B} + \vec{C})}{6} = \frac{3\vec{A} + 3\vec{B} - 2\vec{A} - 2\vec{B} - 2\vec{C}}{6} = \frac{\vec{A} + \vec{B} - 2\vec{C}}{6} \]

7. **Sum the Vectors**:
Now, we sum \( \vec{GD} + \vec{GE} + \vec{GF} \):
\[ \vec{GD} + \vec{GE} + \vec{GF} = \frac{\vec{B} + \vec{C} - 2\vec{A}}{6} + \frac{\vec{A} + \vec{C} - 2\vec{B}}{6} + \frac{\vec{A} + \vec{B} - 2\vec{C}}{6} \]
Combine the terms:
\[ = \frac{(\vec{B} + \vec{C} - 2\vec{A}) + (\vec{A} + \vec{C} - 2\vec{B}) + (\vec{A} + \vec{B} - 2\vec{C})}{6} \]
Simplifying the numerator:
\[ = \frac{0}{6} = \vec{0} \]

### Conclusion:
Thus, we find that:
\[ \vec{GD} + \vec{GE} + \vec{GF} = \vec{0} \]