If the vectors `ahat(i)+hat(j)+hat(k), hat(i)+bhat(j)+hat(k), hat(i)+hat(j)+chat(k)`, where a, b, c are coplanar, then `a+b+c-abc=`

Options

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

Answer

To solve the problem, we need to find the value of \( a + b + c - abc \) 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-by-Step Solution:

1. **Identify the vectors**:
- Let \( \mathbf{A} = \hat{i} + \hat{j} + \hat{k} \)
- Let \( \mathbf{B} = \hat{i} + b\hat{j} + \hat{k} \)
- Let \( \mathbf{C} = \hat{i} + \hat{j} + c\hat{k} \)

2. **Write the vectors in component form**:
- \( \mathbf{A} = (1, 1, 1) \)
- \( \mathbf{B} = (1, b, 1) \)
- \( \mathbf{C} = (1, 1, c) \)

3. **Set up the determinant for coplanarity**:
- The vectors are coplanar if the determinant of the matrix formed by their components is zero:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} = 0 \]

4. **Calculate the determinant**:
- Expanding the determinant:
\[ = 1 \cdot (b \cdot c - 1) - 1 \cdot (1 \cdot c - 1) + 1 \cdot (1 \cdot 1 - b) \]
\[ = bc - 1 - c + 1 + 1 - b \]
\[ = bc - b - c + 1 \]

5. **Set the determinant equal to zero**:
\[ bc - b - c + 1 = 0 \]

6. **Rearranging the equation**:
\[ bc - b - c + 1 = 0 \implies bc - b - c = -1 \]

7. **Express \( a + b + c - abc \)**:
- We need to find \( a + b + c - abc \):
\[ a + b + c = 1 + b + c \]
- From the coplanarity condition, we can express \( abc \):
\[ abc = a \cdot b \cdot c = 1 \cdot b \cdot c \]

8. **Substituting values**:
- Substitute \( a = 1 \):
\[ a + b + c - abc = 1 + b + c - (1 \cdot b \cdot c) \]
\[ = 1 + b + c - bc \]

9. **Using the determinant condition**:
- From the determinant condition \( bc - b - c = -1 \):
\[ 1 + b + c - bc = 1 + (-1) = 2 \]

10. **Final result**:
\[ a + b + c - abc = 2 \]

Thus, the answer is \( \boxed{2} \).