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 will analyze the given vectors and determine which of them is perpendicular to the vector \( \vec{v} \).

### Step 1: Define the vectors
We have the following vectors:
- \( \vec{\alpha} = a \hat{i} + b \hat{j} + c \hat{k} \)
- \( \vec{\beta} = b \hat{i} + c \hat{j} + a \hat{k} \)
- \( \vec{\gamma} = c \hat{i} + a \hat{j} + b \hat{k} \)
- \( \vec{v} = \hat{i} + \hat{j} + \hat{k} \)

### Step 2: Determine coplanarity condition
The vectors \( \vec{\alpha}, \vec{\beta}, \vec{\gamma} \) are coplanar if the scalar triple product is zero. This can be represented using the determinant of a matrix formed by these vectors.

### Step 3: Set up the determinant
We set up the determinant as follows:

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

### Step 4: Calculate the determinant
Calculating the determinant, we expand it:

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

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:

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

### Step 5: Simplify the determinant
Expanding and simplifying gives us:

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

### Step 6: Set the determinant to zero
Since the vectors are coplanar, we set the determinant to zero:

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

### Step 7: Use the identity for cubes
From the identity \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \), we can conclude that \( a + b + c = 0 \) or \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \).

### Step 8: Check perpendicularity with \( \vec{v} \)
To find if \( \vec{v} \) is perpendicular to \( \vec{\alpha}, \vec{\beta}, \vec{\gamma} \), we compute the dot products:

1. \( \vec{v} \cdot \vec{\alpha} = (1)(a) + (1)(b) + (1)(c) = a + b + c \)
2. \( \vec{v} \cdot \vec{\beta} = (1)(b) + (1)(c) + (1)(a) = b + c + a \)
3. \( \vec{v} \cdot \vec{\gamma} = (1)(c) + (1)(a) + (1)(b) = c + a + b \)

Since \( a + b + c = 0 \), we conclude:

\[ \vec{v} \cdot \vec{\alpha} = 0, \quad \vec{v} \cdot \vec{\beta} = 0, \quad \vec{v} \cdot \vec{\gamma} = 0 \]

### Conclusion
Thus, \( \vec{v} \) is perpendicular to all three vectors \( \vec{\alpha}, \vec{\beta}, \vec{\gamma} \).