If the vectors `a hat(i)+hat(j)+hat(k), hat(i)+bhat(j)+hat(k) and hat(i)+hat(j)+chat(k)(a, b, c, !=1)` are coplanar, then the value of `(1)/(1-a)+(1)/(1-b)+(1)/(1-c)` is equal to

Options

A. 0
B. 1
C. a+b+c
D. abc

Explanation

To solve the problem, we need to determine the value of \(\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}\) given that the vectors \(\hat{i} + \hat{j} + \hat{k}\), \(\hat{i} + b\hat{j} + \hat{k}\), and \(\hat{i} + \hat{j} + c\hat{k}\) are coplanar.

### Step 1: Write the vectors
The three vectors can be expressed as:
1. \(\mathbf{v_1} = \hat{i} + \hat{j} + \hat{k}\)
2. \(\mathbf{v_2} = \hat{i} + b\hat{j} + \hat{k}\)
3. \(\mathbf{v_3} = \hat{i} + \hat{j} + c\hat{k}\)

### Step 2: Set up the determinant for coplanarity
For the vectors to be coplanar, the determinant of the matrix formed by these vectors must be equal to zero:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 1 \cdot (b \cdot c - 1) - 1 \cdot (1 \cdot c - 1) + 1 \cdot (1 \cdot 1 - b) \]
This simplifies to:
\[ bc - 1 - c + 1 + 1 - b = bc - b - c + 1 \]
Setting the determinant equal to zero gives:
\[ bc - b - c + 1 = 0 \]
Rearranging, we find:
\[ bc - b - c + 1 = 0 \implies bc - b - c = -1 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ (b-1)(c-1) = 2 \]

### Step 5: Express the required sum
Now we need to find the value of:
\[ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} \]
Using the identity:
\[ \frac{1}{1-x} = \frac{1}{(1-a)(1-b)(1-c)} \cdot (1-b)(1-c) + (1-a)(1-c) + (1-a)(1-b) \]

### Step 6: Substitute values
From the previous steps, we can see that:
\[ \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1 \]

### Conclusion
Thus, the value of \(\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c}\) is equal to \(1\).