If the vectors `phati+hatj+hatk, hati+qhatj+hatk and hati+hatj+rhatk(p!=q!+r!=1)` are coplanar then the value of `pqr-(p+q+r)` is (A) 0 (B) -1 (C) -2 (D) 2

Options

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

Answer

To determine the value of \( pqr - (p + q + r) \) given that the vectors \( \mathbf{a} = p\hat{i} + \hat{j} + \hat{k} \), \( \mathbf{b} = \hat{i} + q\hat{j} + \hat{k} \), and \( \mathbf{c} = \hat{i} + \hat{j} + r\hat{k} \) are coplanar, we can follow these steps:

### Step 1: Set up the vectors
Let:
- \( \mathbf{a} = p\hat{i} + \hat{j} + \hat{k} \)
- \( \mathbf{b} = \hat{i} + q\hat{j} + \hat{k} \)
- \( \mathbf{c} = \hat{i} + \hat{j} + r\hat{k} \)

### Step 2: Use the condition for coplanarity
The vectors are coplanar if the scalar triple product is zero:
\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 0 \]

### Step 3: Compute the cross product \( \mathbf{b} \times \mathbf{c} \)
The cross product \( \mathbf{b} \times \mathbf{c} \) is given by the determinant:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & q & 1 \\ 1 & 1 & r \end{vmatrix} \]

### Step 4: Calculate the determinant
Calculating the determinant:
\[ \mathbf{b} \times \mathbf{c} = \hat{i} \begin{vmatrix} q & 1 \\ 1 & r \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 1 & r \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & q \\ 1 & 1 \end{vmatrix} \]
Calculating each 2x2 determinant:
1. \( \begin{vmatrix} q & 1 \\ 1 & r \end{vmatrix} = qr - 1 \)
2. \( \begin{vmatrix} 1 & 1 \\ 1 & r \end{vmatrix} = r - 1 \)
3. \( \begin{vmatrix} 1 & q \\ 1 & 1 \end{vmatrix} = 1 - q \)

Thus,
\[ \mathbf{b} \times \mathbf{c} = (qr - 1)\hat{i} - (r - 1)\hat{j} + (1 - q)\hat{k} \]

### Step 5: Compute the dot product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \)
Now we compute:
\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (p\hat{i} + \hat{j} + \hat{k}) \cdot ((qr - 1)\hat{i} - (r - 1)\hat{j} + (1 - q)\hat{k}) \]
This results in:
\[ p(qr - 1) + 1(- (r - 1)) + 1(1 - q) \]
Simplifying gives:
\[ p(qr - 1) - (r - 1) + (1 - q) = pqr - p - r + 1 + 1 - q \]
Setting this equal to zero for coplanarity:
\[ pqr - p - q - r + 2 = 0 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ pqr - (p + q + r) = -2 \]

### Final Answer
Thus, the value of \( pqr - (p + q + r) \) is:
\[ \boxed{-2} \]