If ` vec a ,\ vec b ,\ vec c` are three non-zero vectors, no two of which are collinear and the vector ` vec a+ vec b` is collinear with ` vec c ,\ vec b+ vec c` is collinear with ` vec a ,\ t h e n\ vec a+ vec b+ vec c=` ` vec a` b. ` vec b` c. ` vec c` d. none of these

Options

A. `vec(c ) `
B. `vec(0)`
C. `vec(a) + vec(c )`
D. `vec(a )`

Explanation

To solve the problem, we need to analyze the given conditions and derive the expression for the vector sum \( \vec{a} + \vec{b} + \vec{c} \).

### Step-by-Step Solution:

1. **Understand the Conditions**:
We are given two conditions:
- \( \vec{a} + \vec{b} \) is collinear with \( \vec{c} \).
- \( \vec{b} + \vec{c} \) is collinear with \( \vec{a} \).

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

3. **Substituting Expressions**:
We need to find \( \vec{a} + \vec{b} + \vec{c} \). We can substitute the expressions we have:
\[ \vec{a} + \vec{b} + \vec{c} = \lambda \vec{c} + \vec{c} = (\lambda + 1) \vec{c} \]
and from the second condition:
\[ \vec{a} + \vec{b} + \vec{c} = \vec{a} + \mu \vec{a} = (1 + \mu) \vec{a} \]

4. **Equating the Two Expressions**:
Since both expressions represent \( \vec{a} + \vec{b} + \vec{c} \), we can set them equal to each other:
\[ (\lambda + 1) \vec{c} = (1 + \mu) \vec{a} \]

5. **Comparing Coefficients**:
For the above equation to hold true, the coefficients of \( \vec{a} \) and \( \vec{c} \) must be equal:
- Coefficient of \( \vec{c} \): \( \lambda + 1 = 0 \) → \( \lambda = -1 \)
- Coefficient of \( \vec{a} \): \( 1 + \mu = 0 \) → \( \mu = -1 \)

6. **Substituting Back**:
Now substituting \( \lambda \) and \( \mu \) back into the expressions for \( \vec{a} + \vec{b} \):
- From \( \vec{a} + \vec{b} = \lambda \vec{c} \):
\[ \vec{a} + \vec{b} = -1 \cdot \vec{c} = -\vec{c} \]
- From \( \vec{b} + \vec{c} = \mu \vec{a} \):
\[ \vec{b} + \vec{c} = -1 \cdot \vec{a} = -\vec{a} \]

7. **Final Expression**:
Now substituting back into \( \vec{a} + \vec{b} + \vec{c} \):
\[ \vec{a} + \vec{b} + \vec{c} = \vec{a} + (-\vec{c}) + \vec{c} = \vec{0} \]

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

### Answer:
The answer is **(d) none of these**, as the sum is the zero vector.