If the vectors `ahati+hatj+hatk, hati+bhatj+hatk, hati+hatj+chatk(a!=1, b!=1,c!=1)` are coplanar then the value of `1/(1-a)+1/(1-b)+1/(1-c)` is (A) 0 (B) 1 (C) -1 (D) 2

Explanation

To solve the problem, we need to determine the condition under which the given vectors are coplanar and then find the value of \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \).

### Step 1: Write the vectors in matrix form
The vectors given are:
1. \( \mathbf{v_1} = a\hat{i} + \hat{j} + \hat{k} \)
2. \( \mathbf{v_2} = \hat{i} + b\hat{j} + \hat{k} \)
3. \( \mathbf{v_3} = \hat{i} + \hat{j} + c\hat{k} \)

We can represent these vectors as a matrix:
\[ \begin{bmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{bmatrix} \]

### Step 2: Set up the determinant condition for coplanarity
For the vectors to be coplanar, the determinant of this matrix must be zero:
\[ \text{Det} = \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Using the determinant formula for a \( 3 \times 3 \) matrix, we expand:
\[ \text{Det} = a \begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} + 1 \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} \]

Calculating the smaller determinants:
1. \( \begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} = bc - 1 \)
2. \( \begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} = c - 1 \)
3. \( \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} = 1 - b \)

Substituting these back into the determinant:
\[ \text{Det} = a(bc - 1) - (c - 1) + (1 - b) \]
\[ = abc - a - c + 1 + 1 - b \]
\[ = abc - a - b - c + 2 \]

Setting the determinant to zero gives:
\[ abc - a - b - c + 2 = 0 \]
\[ abc = a + b + c - 2 \]

### Step 4: Rearranging the equation
We can rearrange this equation to express \( abc \):
\[ abc - a - b - c + 2 = 0 \]

### Step 5: Find the value of \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \)
Using the relationship obtained from the determinant, we can manipulate the expression:
\[ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = \frac{(1-b)(1-c) + (1-a)(1-c) + (1-a)(1-b)}{(1-a)(1-b)(1-c)} \]

Expanding the numerator:
\[ = (1 - b - c + bc) + (1 - a - c + ac) + (1 - a - b + ab) \]
\[ = 3 - (a + b + c) + (ab + ac + bc) \]

Using the condition \( abc = a + b + c - 2 \), we can substitute \( a + b + c \) and \( ab + ac + bc \) back into our equation.

After simplification, we find:
\[ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1 \]

### Final Answer
Thus, the value of \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \) is \( \boxed{1} \).