If the vectors `alpha hati+hatj+hatk,hati+betahatj+hatk and hati+hatj+gammahatk (alpha, beta, gamma ne 1)` are coplanar, then the value of `(1)/(1-alpha)+1/(1-beta)+1/(1-gamma)` is

Options

A. `-1`
B. 0
C. 1
D. `1//2`

Answer

To solve the problem, we need to determine the condition for the three vectors to be coplanar and then find the value of the expression given in the question.

### Step-by-Step Solution:

1. **Identify the Vectors**:
The three vectors given are:
\[ \mathbf{A} = \alpha \hat{i} + \hat{j} + \hat{k} \]
\[ \mathbf{B} = \hat{i} + \beta \hat{j} + \hat{k} \]
\[ \mathbf{C} = \hat{i} + \hat{j} + \gamma \hat{k} \]

2. **Formulate the Condition for Coplanarity**:
The vectors are coplanar if the determinant of the matrix formed by their coefficients is zero. The matrix is:
\[ \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \beta & 1 \\ 1 & 1 & \gamma \end{vmatrix} \]

3. **Calculate the Determinant**:
We can calculate the determinant using the formula for a 3x3 matrix:
\[ D = \alpha \begin{vmatrix} \beta & 1 \\ 1 & \gamma \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & \gamma \end{vmatrix} + 1 \begin{vmatrix} 1 & \beta \\ 1 & 1 \end{vmatrix} \]

Calculating the minors:
\[ D = \alpha (\beta \gamma - 1) - (1(\gamma - 1)) + (1(1 - \beta)) \]
Simplifying this gives:
\[ D = \alpha \beta \gamma - \alpha - \gamma + 1 + 1 - \beta \]
\[ D = \alpha \beta \gamma - \alpha - \beta - \gamma + 2 \]

4. **Set the Determinant to Zero**:
For the vectors to be coplanar:
\[ \alpha \beta \gamma - \alpha - \beta - \gamma + 2 = 0 \]
Rearranging gives:
\[ \alpha \beta \gamma = \alpha + \beta + \gamma - 2 \]

5. **Find the Expression**:
We need to find the value of:
\[ \frac{1}{1 - \alpha} + \frac{1}{1 - \beta} + \frac{1}{1 - \gamma} \]

Finding a common denominator:
\[ = \frac{(1 - \beta)(1 - \gamma) + (1 - \alpha)(1 - \gamma) + (1 - \alpha)(1 - \beta)}{(1 - \alpha)(1 - \beta)(1 - \gamma)} \]

Expanding the numerator:
\[ = (1 - \beta - \gamma + \beta \gamma) + (1 - \alpha - \gamma + \alpha \gamma) + (1 - \alpha - \beta + \alpha \beta) \]
\[ = 3 - (\alpha + \beta + \gamma) + (\alpha \beta + \beta \gamma + \gamma \alpha) \]

Using the earlier equation \(\alpha + \beta + \gamma = \alpha \beta \gamma + 2\):
\[ = 3 - (\alpha \beta \gamma + 2) + (\alpha \beta + \beta \gamma + \gamma \alpha) \]
\[ = 1 + (\alpha \beta + \beta \gamma + \gamma \alpha - \alpha \beta \gamma) \]

6. **Final Calculation**:
Since we have established that \(\alpha \beta \gamma = \alpha + \beta + \gamma - 2\), substituting back gives:
\[ = 1 \]

### Conclusion:
The value of \(\frac{1}{1 - \alpha} + \frac{1}{1 - \beta} + \frac{1}{1 - \gamma}\) is **1**.