Let `hata,hatb` and `hatc` be unit vectors and `alpha, beta`, and `gamma` the angles between the vectors `hata,hatb,hatb,hatc` and `hatc, hata` respectively . If `hata + hatb+hatc` is also a unit vectors, then `cosalpha + cosbeta +cosgamma` is equal to -

Options

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

Explanation

To solve the problem, we need to find the value of \( \cos \alpha + \cos \beta + \cos \gamma \) given that \( \hat{a}, \hat{b}, \hat{c} \) are unit vectors and \( \hat{a} + \hat{b} + \hat{c} \) is also a unit vector.

### Step-by-step Solution:

1. **Understanding the Unit Vectors**:
Since \( \hat{a}, \hat{b}, \hat{c} \) are unit vectors, we have:
\[ |\hat{a}| = 1, \quad |\hat{b}| = 1, \quad |\hat{c}| = 1 \]

2. **Using the Given Condition**:
We know that \( \hat{a} + \hat{b} + \hat{c} \) is also a unit vector:
\[ |\hat{a} + \hat{b} + \hat{c}| = 1 \]

3. **Squaring Both Sides**:
Squaring the magnitude of the sum gives:
\[ |\hat{a} + \hat{b} + \hat{c}|^2 = 1^2 \]
Expanding the left side:
\[ (\hat{a} + \hat{b} + \hat{c}) \cdot (\hat{a} + \hat{b} + \hat{c}) = 1 \]
This expands to:
\[ \hat{a} \cdot \hat{a} + \hat{b} \cdot \hat{b} + \hat{c} \cdot \hat{c} + 2(\hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{c} + \hat{c} \cdot \hat{a}) = 1 \]

4. **Substituting the Values**:
Since each vector is a unit vector, we have:
\[ 1 + 1 + 1 + 2(\hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{c} + \hat{c} \cdot \hat{a}) = 1 \]
This simplifies to:
\[ 3 + 2(\hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{c} + \hat{c} \cdot \hat{a}) = 1 \]

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

6. **Relating to Cosines**:
The dot products can be expressed in terms of the angles:
\[ \hat{a} \cdot \hat{b} = \cos \alpha, \quad \hat{b} \cdot \hat{c} = \cos \beta, \quad \hat{c} \cdot \hat{a} = \cos \gamma \]
Thus, we have:
\[ \cos \alpha + \cos \beta + \cos \gamma = -1 \]

### Final Answer:
\[ \cos \alpha + \cos \beta + \cos \gamma = -1 \]