If the vectots `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

Options

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

Explanation

To solve the problem, we need 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.

### Step 1: Set up the condition for coplanarity
Vectors are coplanar if the scalar triple product is zero. The scalar triple product can be represented as the determinant of a matrix formed by the vectors:

\[ \begin{vmatrix} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant
Now we will compute the determinant:

\[ \begin{vmatrix} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r \end{vmatrix} \]

Expanding this determinant along the first row gives:

\[ = p \begin{vmatrix} q & 1 \\ 1 & r \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & r \end{vmatrix} + 1 \begin{vmatrix} 1 & q \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} q & 1 \\ 1 & r \end{vmatrix} = qr - 1 \)
2. \( \begin{vmatrix} 1 & 1 \\ 1 & r \end{vmatrix} = r - 1 \)
3. \( \begin{vmatrix} 1 & q \\ 1 & 1 \end{vmatrix} = 1 - q \)

Substituting these back into the determinant:

\[ = p(qr - 1) - (r - 1) + (1 - q) \]

### Step 3: Simplify the expression
Now we simplify the expression:

\[ = pqr - p - r + 1 + 1 - q \]
\[ = pqr - p - q - r + 2 \]

### Step 4: Set the determinant to zero
Since the vectors are coplanar, we set the determinant to zero:

\[ pqr - p - q - r + 2 = 0 \]

### Step 5: Solve for \( pqr - (p + q + r) \)
Rearranging gives us:

\[ pqr - (p + q + r) = -2 \]

Thus, the value of \( pqr - (p + q + r) \) is:

\[ \boxed{-2} \]