Let `vec(alpha)=ahati+bhatj+chatk, vec(beta)=bhati+chatj+ahatk` and `vec(gamma)=chati+ahatj+bhatk` be three coplnar vectors with `a!=b`, and `vecv=hati+hatj+hatk`. Then `vecv` is perpendicular to

Options

A. `vecalpha`
B. `vecbeta`
C. `vecgamma`
D. none of these

Explanation

To solve the problem step by step, we need to determine which vectors the vector \( \vec{v} = \hat{i} + \hat{j} + \hat{k} \) is perpendicular to among the given coplanar vectors \( \vec{\alpha}, \vec{\beta}, \) and \( \vec{\gamma} \).

### Step 1: Write down the vectors
Given:
- \( \vec{\alpha} = a \hat{i} + b \hat{j} + c \hat{k} \)
- \( \vec{\beta} = b \hat{i} + c \hat{j} + a \hat{k} \)
- \( \vec{\gamma} = c \hat{i} + a \hat{j} + b \hat{k} \)
- \( \vec{v} = \hat{i} + \hat{j} + \hat{k} \)

### Step 2: Check for coplanarity
The vectors \( \vec{\alpha}, \vec{\beta}, \) and \( \vec{\gamma} \) are coplanar if the determinant of the matrix formed by these vectors is zero. The determinant is given by:

\[ \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant:

\[ = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. \( \begin{vmatrix} c & a \\ a & b \end{vmatrix} = cb - a^2 \)
2. \( \begin{vmatrix} b & a \\ c & b \end{vmatrix} = bb - ac = b^2 - ac \)
3. \( \begin{vmatrix} b & c \\ c & a \end{vmatrix} = ba - c^2 \)

Putting it all together:

\[ = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) \]

Expanding this gives:

\[ = acb - a^3 - b^3 + abc + abc - c^3 \]

Combining like terms yields:

\[ = 3abc - (a^3 + b^3 + c^3) = 0 \]

### Step 4: Use the condition \( a + b + c = 0 \)
From the determinant condition, we can derive that:

\[ a + b + c = 0 \]

### Step 5: Check perpendicularity with \( \vec{v} \)
To check if \( \vec{v} \) is perpendicular to \( \vec{\alpha}, \vec{\beta}, \) and \( \vec{\gamma} \), we compute the dot products:

1. **Dot product with \( \vec{\alpha} \)**:
\[ \vec{v} \cdot \vec{\alpha} = (\hat{i} + \hat{j} + \hat{k}) \cdot (a \hat{i} + b \hat{j} + c \hat{k}) = a + b + c = 0 \]

2. **Dot product with \( \vec{\beta} \)**:
\[ \vec{v} \cdot \vec{\beta} = (\hat{i} + \hat{j} + \hat{k}) \cdot (b \hat{i} + c \hat{j} + a \hat{k}) = b + c + a = 0 \]

3. **Dot product with \( \vec{\gamma} \)**:
\[ \vec{v} \cdot \vec{\gamma} = (\hat{i} + \hat{j} + \hat{k}) \cdot (c \hat{i} + a \hat{j} + b \hat{k}) = c + a + b = 0 \]

### Conclusion
Since \( \vec{v} \) is perpendicular to \( \vec{\alpha}, \vec{\beta}, \) and \( \vec{\gamma} \):

\[ \vec{v} \text{ is perpendicular to } \vec{\alpha}, \vec{\beta}, \text{ and } \vec{\gamma}. \]