If the vectors ai + j + k , i + bj + k and i + j + ck, where a, b, c `ne1`, are coplanar,
then : `1/(1-a)+1/(1-b)+1/(1-c)=…`

Options

A. `-1`
B. 0
C. 1
D. 3

Answer

To solve the problem, we need to determine the condition for the vectors \( \mathbf{a} = ai + j + k \), \( \mathbf{b} = i + bj + k \), and \( \mathbf{c} = i + j + ck \) to be coplanar. The condition for coplanarity of three vectors is that their scalar triple product is zero.

### Step-by-Step Solution:

1. **Write the Vectors:**
The vectors are given as:
\[ \mathbf{A} = ai + j + k, \quad \mathbf{B} = i + bj + k, \quad \mathbf{C} = i + j + ck \]

2. **Set Up the Scalar Triple Product:**
The scalar triple product can be computed using the determinant of a matrix formed by the coefficients of the vectors:
\[ \text{Scalar Triple Product} = \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} \]

3. **Calculate the Determinant:**
We can calculate the determinant using the formula for a 3x3 matrix:
\[ \text{Determinant} = a \begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} + 1 \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} \]
This expands to:
\[ = a(bc - 1) - (c - 1) + (1 - b) = abc - a - c + 1 + 1 - b \]
Simplifying gives:
\[ abc - a - b - c + 2 \]

4. **Set the Determinant Equal to Zero:**
For the vectors to be coplanar, we set the determinant equal to zero:
\[ abc - a - b - c + 2 = 0 \]

5. **Rearranging the Equation:**
Rearranging gives:
\[ abc = a + b + c - 2 \]

6. **Substituting Values:**
Now we need to find \( \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \). We can substitute \( a = -2 \), \( b = -2 \), and \( c = -2 \) (as derived from the determinant conditions):
\[ \frac{1}{1 - (-2)} + \frac{1}{1 - (-2)} + \frac{1}{1 - (-2)} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \]

### Final Result:
Thus, we conclude that:
\[ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1 \]