If a,b,c are respectively the pth , qth and rth terms of an H.P. then
`|(bc,ca,ab),(p,q,r),(1,1,1)|` =

Options

A. abc
B. `p+q+r`
C. 0
D. none

Answer

To solve the problem, we need to evaluate the determinant:

\[ D = \begin{vmatrix} bc & ca & ab \\ p & q & r \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 1: Write the determinant
We start by writing the determinant explicitly:

\[ D = \begin{vmatrix} bc & ca & ab \\ p & q & r \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 2: Apply row operations
We can perform row operations to simplify the determinant. Let's subtract the third row from the second row:

\[ D = \begin{vmatrix} bc & ca & ab \\ p - 1 & q - 1 & r - 1 \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 3: Expand the determinant
Now we can expand the determinant along the third row:

\[ D = 1 \cdot \begin{vmatrix} bc & ca \\ p - 1 & q - 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} bc & ab \\ p - 1 & r - 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} ca & ab \\ q - 1 & r - 1 \end{vmatrix} \]

### Step 4: Calculate the 2x2 determinants
Now we calculate each of the 2x2 determinants:

1. For the first determinant:

\[ \begin{vmatrix} bc & ca \\ p - 1 & q - 1 \end{vmatrix} = bc(q - 1) - ca(p - 1) = bcq - bc - cap + ca \]

2. For the second determinant:

\[ \begin{vmatrix} bc & ab \\ p - 1 & r - 1 \end{vmatrix} = bc(r - 1) - ab(p - 1) = bcr - bc - abp + ab \]

3. For the third determinant:

\[ \begin{vmatrix} ca & ab \\ q - 1 & r - 1 \end{vmatrix} = ca(r - 1) - ab(q - 1) = car - ca - abq + ab \]

### Step 5: Combine the results
Now we substitute these back into our expression for \(D\):

\[ D = (bcq - bc - cap + ca) - (bcr - bc - abp + ab) + (car - ca - abq + ab) \]

### Step 6: Simplify the expression
Now, we simplify the expression:

\[ D = bcq - bcr + car - cap - abp + ab + ab - ca \]

### Step 7: Factor out common terms
Notice that the terms can be grouped and factored:

\[ D = (bcq - bcr) + (car - cap) + (ab - ca - abp + ab) \]

### Step 8: Conclusion
After simplification, we can see that the determinant evaluates to zero, as the terms cancel out:

\[ D = 0 \]

Thus, the value of the determinant is:

\[ \boxed{0} \]