Let `vec(a), vec(b), vec(c)` are three unit vector such that `vec(a)+vec(b)+vec(c)` os also a unit vector. If pairwise angles between `vec(a), vec(b), vec(c)` are `theta_(1), theta_(2)` and `theta_(3)` respectively then `cos theta_(1)+cos theta_(2)+ cos theta_(3)` equals

Options

A. `3`
B. `-3`
C. `1`
D. `-1`

Explanation

To solve the problem, we need to find the value of \( \cos \theta_1 + \cos \theta_2 + \cos \theta_3 \) given that \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors and \( \vec{a} + \vec{b} + \vec{c} \) is also a unit vector.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have three unit vectors \( \vec{a}, \vec{b}, \vec{c} \) such that:
\[ |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \]
- The sum of these vectors is also a unit vector:
\[ |\vec{a} + \vec{b} + \vec{c}| = 1 \]

2. **Using the Magnitude Condition**:
- The magnitude of the sum can be expressed as:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) \]
- Expanding this, we get:
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \]
- Since \( |\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1 \):
\[ 1 + 1 + 1 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 1 \]
- This simplifies to:
\[ 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 1 \]

3. **Setting Up the Equation**:
- Rearranging gives:
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 1 - 3 \]
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -2 \]
- Dividing by 2:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -1 \]

4. **Relating Dot Products to Cosines**:
- The dot products can be expressed in terms of the angles:
\[ \vec{a} \cdot \vec{b} = \cos \theta_1, \quad \vec{b} \cdot \vec{c} = \cos \theta_2, \quad \vec{c} \cdot \vec{a} = \cos \theta_3 \]
- Thus, we have:
\[ \cos \theta_1 + \cos \theta_2 + \cos \theta_3 = -1 \]

5. **Final Result**:
- Therefore, the value of \( \cos \theta_1 + \cos \theta_2 + \cos \theta_3 \) is:
\[ \boxed{-1} \]