If ` vec a ,\ vec b ,\ vec c` are three non-zero vectors, no two which are collinear and the vector ` vec a+ vec b` is collinear with ` vec c ,\ vec b+ vec c` is collinear with ` vec a` then, `vec a+ vec b+ vec c=`

Options

A. a) `vec a`
B. b) ` vec b`
C. c) ` vec c`
D. d) None of these

Explanation

To solve the problem, we need to analyze the given conditions about the vectors \( \vec{a}, \vec{b}, \vec{c} \).

### Step 1: Understand the conditions
We know that:
1. \( \vec{a} + \vec{b} \) is collinear with \( \vec{c} \).
2. \( \vec{b} + \vec{c} \) is collinear with \( \vec{a} \).

### Step 2: Express the collinearity conditions
From the first condition, we can express it mathematically as:
\[ \vec{a} + \vec{b} = k_1 \vec{c} \quad \text{(1)} \]
for some scalar \( k_1 \).

From the second condition, we can express it as:
\[ \vec{b} + \vec{c} = k_2 \vec{a} \quad \text{(2)} \]
for some scalar \( k_2 \).

### Step 3: Rearranging the equations
From equation (1), we can rearrange it to find \( \vec{b} \):
\[ \vec{b} = k_1 \vec{c} - \vec{a} \quad \text{(3)} \]

From equation (2), we can rearrange it to find \( \vec{c} \):
\[ \vec{c} = k_2 \vec{a} - \vec{b} \quad \text{(4)} \]

### Step 4: Substitute equation (3) into equation (4)
Substituting equation (3) into equation (4):
\[ \vec{c} = k_2 \vec{a} - (k_1 \vec{c} - \vec{a}) \]
This simplifies to:
\[ \vec{c} + k_1 \vec{c} = k_2 \vec{a} + \vec{a} \]
\[ (1 + k_1) \vec{c} = (k_2 + 1) \vec{a} \]

### Step 5: Analyze the equation
Since \( \vec{a} \) and \( \vec{c} \) are not collinear (given in the problem), the coefficients must equal zero:
\[ 1 + k_1 = 0 \quad \text{and} \quad k_2 + 1 = 0 \]
Thus, we find:
\[ k_1 = -1 \quad \text{and} \quad k_2 = -1 \]

### Step 6: Substitute back to find relationships
Substituting \( k_1 \) and \( k_2 \) back into equations (1) and (2):
From equation (1):
\[ \vec{a} + \vec{b} = -\vec{c} \quad \text{(5)} \]

From equation (2):
\[ \vec{b} + \vec{c} = -\vec{a} \quad \text{(6)} \]

### Step 7: Add equations (5) and (6)
Now, we can add equations (5) and (6):
\[ (\vec{a} + \vec{b}) + (\vec{b} + \vec{c}) = -\vec{c} - \vec{a} \]
This simplifies to:
\[ \vec{a} + 2\vec{b} + \vec{c} = -\vec{c} - \vec{a} \]
Rearranging gives:
\[ \vec{a} + 2\vec{b} + \vec{c} + \vec{c} + \vec{a} = 0 \]
\[ 2\vec{a} + 2\vec{b} + 2\vec{c} = 0 \]
Dividing by 2:
\[ \vec{a} + \vec{b} + \vec{c} = 0 \]

### Conclusion
Thus, we conclude:
\[ \vec{a} + \vec{b} + \vec{c} = 0 \]