`veca,vecb` and `vecc` are three non-zero vectors, no two of which are collinear and the vectors `veca+vecb` is collinear with `vecc`, `vecb+vecc` is collinear with `veca`, then `veca+vecb+vecc=`

Options

A. `vec a`
B. `vec b`
C. `vec c`
D. none of these

Explanation

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

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

From the definition of collinearity, we can express these conditions mathematically:
- There exists a scalar \( \lambda \) such that:
\[ \vec{a} + \vec{b} = \lambda \vec{c} \]
- There exists a scalar \( \mu \) such that:
\[ \vec{b} + \vec{c} = \mu \vec{a} \]

### Step 2: Rearranging the equations
From the second equation, we can express \( \vec{c} \) in terms of \( \vec{a} \) and \( \vec{b} \):
\[ \vec{c} = \frac{1}{\mu} (\vec{b} + \vec{c}) \implies \vec{c} = \mu \vec{a} - \vec{b} \]

### Step 3: Substitute \( \vec{c} \) into the first equation
Substituting \( \vec{c} \) from the above expression into the first equation:
\[ \vec{a} + \vec{b} = \lambda (\mu \vec{a} - \vec{b}) \]
Distributing \( \lambda \):
\[ \vec{a} + \vec{b} = \lambda \mu \vec{a} - \lambda \vec{b} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \vec{a} + \vec{b} + \lambda \vec{b} = \lambda \mu \vec{a} \]
This can be rewritten as:
\[ \vec{a} + (1 + \lambda) \vec{b} = \lambda \mu \vec{a} \]

### Step 5: Comparing coefficients
Now, we can compare the coefficients of \( \vec{a} \) and \( \vec{b} \):
1. Coefficient of \( \vec{a} \): \( 1 = \lambda \mu \)
2. Coefficient of \( \vec{b} \): \( 1 + \lambda = 0 \)

From \( 1 + \lambda = 0 \), we find:
\[ \lambda = -1 \]

Substituting \( \lambda = -1 \) into \( 1 = \lambda \mu \):
\[ 1 = -\mu \implies \mu = -1 \]

### Step 6: Substitute back to find \( \vec{c} \)
Now substituting \( \lambda \) back into the expression for \( \vec{c} \):
\[ \vec{a} + \vec{b} = -\vec{c} \implies \vec{a} + \vec{b} + \vec{c} = 0 \]

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