`' 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. 0
B. (a+b+c)BC
C. (a+b+c)AC
D. (a+b+c)AB

Explanation

To solve the problem, we need to show that \( a \vec{I}A + b \vec{I}B + c \vec{I}C = \vec{0} \), 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. It is also the center of the circle inscribed in the triangle.

2. **Position Vectors**:
Let the position vectors of points \( A, B, C \) be denoted as \( \vec{A}, \vec{B}, \vec{C} \). The position vector of the incenter \( I \) can be expressed as:
\[ \vec{I} = \frac{a \vec{A} + b \vec{B} + c \vec{C}}{a + b + c} \]

3. **Expressing Vectors \( \vec{IA}, \vec{IB}, \vec{IC} \)**:
The vectors from the incenter to the vertices can be expressed as:
\[ \vec{IA} = \vec{A} - \vec{I}, \quad \vec{IB} = \vec{B} - \vec{I}, \quad \vec{IC} = \vec{C} - \vec{I} \]

4. **Substituting the Incenter**:
Substitute the expression for \( \vec{I} \) into the equations for \( \vec{IA}, \vec{IB}, \vec{IC} \):
\[ \vec{IA} = \vec{A} - \frac{a \vec{A} + b \vec{B} + c \vec{C}}{a + b + c} \]
\[ \vec{IB} = \vec{B} - \frac{a \vec{A} + b \vec{B} + c \vec{C}}{a + b + c} \]
\[ \vec{IC} = \vec{C} - \frac{a \vec{A} + b \vec{B} + c \vec{C}}{a + b + c} \]

5. **Calculating \( a \vec{IA} + b \vec{IB} + c \vec{IC} \)**:
Now, we can compute:
\[ a \vec{IA} + b \vec{IB} + c \vec{IC} = a \left( \vec{A} - \vec{I} \right) + b \left( \vec{B} - \vec{I} \right) + c \left( \vec{C} - \vec{I} \right) \]
\[ = a \vec{A} + b \vec{B} + c \vec{C} - (a + b + c) \vec{I} \]

6. **Substituting for \( \vec{I} \)**:
Substitute \( \vec{I} \) back into the equation:
\[ = a \vec{A} + b \vec{B} + c \vec{C} - (a + b + c) \left( \frac{a \vec{A} + b \vec{B} + c \vec{C}}{a + b + c} \right) \]
\[ = a \vec{A} + b \vec{B} + c \vec{C} - (a \vec{A} + b \vec{B} + c \vec{C}) = \vec{0} \]

7. **Conclusion**:
Therefore, we conclude that:
\[ a \vec{IA} + b \vec{IB} + c \vec{IC} = \vec{0} \]

### Final Answer:
Thus, the correct option is **(a) \( \vec{0} \)**.