`' I '` is the incentre of triangle `A B C` whose corresponding sides are `a , b ,c ,` rspectively. `a vec I A+b vec I B+c vec I C` is always equal to a. ` vec0` b. `(a+b+c) vec B C` c. `( vec a+ vec b+ vec c) vec A C` d. `(a+b+c) vec A B`

Options

A. `vec0`
B. `(a+ b + c) vec(BC)`
C. `(veca+ vecb+vecc) vec(AC)`
D. `(a+b+c) vec(AB)`

Explanation

To solve the problem, we need to find the value of the expression \( a \vec{IA} + b \vec{IB} + c \vec{IC} \), where \( I \) is the incenter of triangle \( ABC \) and \( a, b, c \) are the lengths of the sides opposite to vertices \( A, B, C \) respectively.

### Step-by-Step Solution:

1. **Understanding the Incenter**:
The incenter \( I \) of triangle \( ABC \) is the point where the angle bisectors of the triangle intersect, and it is equidistant from all three sides of the triangle.

2. **Position Vectors**:
Let us denote the position vectors of points \( A, B, C \) as \( \vec{P}, \vec{Q}, \vec{R} \) respectively. Therefore:
- \( \vec{IA} = \vec{P} - \vec{I} \)
- \( \vec{IB} = \vec{Q} - \vec{I} \)
- \( \vec{IC} = \vec{R} - \vec{I} \)

3. **Assuming the Incenter as Origin**:
For simplification, we can assume the position vector of the incenter \( I \) is the origin. Thus, \( \vec{I} = \vec{0} \). This leads to:
- \( \vec{IA} = \vec{P} \)
- \( \vec{IB} = \vec{Q} \)
- \( \vec{IC} = \vec{R} \)

4. **Using the Formula for Incenter**:
The position vector of the incenter \( I \) can also be expressed as:
\[ \vec{I} = \frac{a \vec{A} + b \vec{B} + c \vec{C}}{a + b + c} \]
Since we assumed \( \vec{I} = \vec{0} \), we have:
\[ \vec{0} = \frac{a \vec{P} + b \vec{Q} + c \vec{R}}{a + b + c} \]

5. **Cross Multiplying**:
By cross-multiplying, we get:
\[ a \vec{P} + b \vec{Q} + c \vec{R} = \vec{0} \]

6. **Final Expression**:
Thus, we can conclude that:
\[ a \vec{IA} + b \vec{IB} + c \vec{IC} = \vec{0} \]

### Conclusion:
The value of \( a \vec{IA} + b \vec{IB} + c \vec{IC} \) is the zero vector \( \vec{0} \).

### Answer:
The correct option is **(a) \( \vec{0} \)**.