Let a,b,c are three vectors of which every pair is non-collinear, if the vectors a+b and b+c are collinear with c annd a respectively, then find a+b+c.

Answer

To solve the problem, we need to find the vector \( a + b + c \) given the conditions about the vectors \( a \), \( b \), and \( c \). Let's break down the solution step by step.

### Step 1: Set up the equations based on collinearity
We have two conditions given in the problem:
1. The vector \( a + b \) is collinear with \( c \).
2. The vector \( b + c \) is collinear with \( a \).

From the first condition, we can express this as:
\[ a + b = k \cdot c \quad \text{(for some scalar } k\text{)} \]
This is our **Equation (1)**.

From the second condition, we can express this as:
\[ b + c = \mu \cdot a \quad \text{(for some scalar } \mu\text{)} \]
This is our **Equation (2)**.

### Step 2: Substitute and manipulate the equations
Now, we want to find \( a + b + c \). We can express \( a + b + c \) using the equations we have.

From **Equation (1)**, we can express \( a + b \) as:
\[ a + b + c = k \cdot c + c = (k + 1) \cdot c \]
This gives us **Equation (3)**.

From **Equation (2)**, we can express \( b + c \) as:
\[ a + b + c = a + \mu \cdot a = (1 + \mu) \cdot a \]
This gives us **Equation (4)**.

### Step 3: Set the two expressions for \( a + b + c \) equal to each other
Now we have two expressions for \( a + b + c \):
1. From **Equation (3)**: \( a + b + c = (k + 1) \cdot c \)
2. From **Equation (4)**: \( a + b + c = (1 + \mu) \cdot a \)

Since both expressions are equal, we can set them equal to each other:
\[ (k + 1) \cdot c = (1 + \mu) \cdot a \]

### Step 4: Analyze the relationship between \( a \) and \( c \)
Given that \( a \) and \( c \) are non-collinear vectors, the only way for the above equation to hold true is if both sides equal zero. This leads us to conclude:
\[ k + 1 = 0 \quad \text{and} \quad 1 + \mu = 0 \]
Thus, we find:
\[ k = -1 \quad \text{and} \quad \mu = -1 \]

### Step 5: Substitute back to find \( a + b + c \)
Now substituting \( k \) and \( \mu \) back into either equation, we find:
\[ a + b + c = (k + 1) \cdot c = 0 \cdot c = 0 \]
or
\[ a + b + c = (1 + \mu) \cdot a = 0 \cdot a = 0 \]

### Conclusion
Thus, the final result is:
\[ \boxed{0} \]