If `vec a , vec b , vec c ` are three non-zero vectors (no two of which are collinear), such that the pairs of vectors `(vec a + vec b, vec c) and (vec b + vec c , vec a) ` are collinear, then `vec a + vec b + vec c =`

Options

A. `vec a`
B. `vec b`
C. ` vec c `
D. `vec 0`

Answer

To solve the problem, we need to analyze the conditions given about the vectors \( \vec{a}, \vec{b}, \vec{c} \). We know that the pairs of vectors \( (\vec{a} + \vec{b}, \vec{c}) \) and \( (\vec{b} + \vec{c}, \vec{a}) \) are collinear.

### Step-by-Step Solution:

1. **Understanding Collinearity**:
- Two vectors \( \vec{u} \) and \( \vec{v} \) are collinear if there exists a scalar \( \lambda \) such that \( \vec{u} = \lambda \vec{v} \).
- For our case:
- \( \vec{a} + \vec{b} = \lambda_1 \vec{c} \) (1)
- \( \vec{b} + \vec{c} = \lambda_2 \vec{a} \) (2)

2. **Rearranging Equation (1)**:
- From equation (1), we can express \( \vec{a} \):
\[ \vec{a} = \lambda_1 \vec{c} - \vec{b} \tag{3} \]

3. **Substituting Equation (3) into Equation (2)**:
- Substitute \( \vec{a} \) from equation (3) into equation (2):
\[ \vec{b} + \vec{c} = \lambda_2 (\lambda_1 \vec{c} - \vec{b}) \]
- Expanding this gives:
\[ \vec{b} + \vec{c} = \lambda_2 \lambda_1 \vec{c} - \lambda_2 \vec{b} \]

4. **Rearranging the Equation**:
- Bring all terms involving \( \vec{b} \) to one side:
\[ \vec{b} + \lambda_2 \vec{b} = \lambda_2 \lambda_1 \vec{c} - \vec{c} \]
- Factor out \( \vec{b} \):
\[ (1 + \lambda_2) \vec{b} = (\lambda_2 \lambda_1 - 1) \vec{c} \tag{4} \]

5. **Setting Coefficients to Zero**:
- Since \( \vec{b} \) and \( \vec{c} \) are non-collinear, the coefficients of \( \vec{b} \) and \( \vec{c} \) must independently equal zero:
- From equation (4):
\[ 1 + \lambda_2 = 0 \implies \lambda_2 = -1 \]
\[ \lambda_2 \lambda_1 - 1 = 0 \implies -\lambda_1 - 1 = 0 \implies \lambda_1 = -1 \]

6. **Finding \( \vec{a} + \vec{b} + \vec{c} \)**:
- Substitute \( \lambda_1 \) and \( \lambda_2 \) back into equation (1):
\[ \vec{a} + \vec{b} = -1 \cdot \vec{c} \implies \vec{a} + \vec{b} + \vec{c} = 0 \]

### Final Result:
\[ \vec{a} + \vec{b} + \vec{c} = \vec{0} \]