If a,b,c be respectively the `p^(th),q^(th)andr^(th)` terms of a H.P., then
`Delta=|{:(bc,ca,ab),(p,q,r),(1,1,1):}|` equals

Options

A. 1
B. 0
C. -1
D. pqr

Answer

To solve the problem, we need to find the value of the determinant \( \Delta = \begin{vmatrix} bc & ca & ab \\ p & q & r \\ 1 & 1 & 1 \end{vmatrix} \).

### Step 1: Express terms of H.P. in terms of A.P.
Since \( a, b, c \) are the \( p^{th}, q^{th}, r^{th} \) terms of a Harmonic Progression (H.P.), we can express them in terms of their corresponding terms in an Arithmetic Progression (A.P.).

The terms of H.P. can be expressed as:
\[ \frac{1}{a} = A + (p-1)D \]
\[ \frac{1}{b} = A + (q-1)D \]
\[ \frac{1}{c} = A + (r-1)D \]
where \( A \) is the first term of the A.P. and \( D \) is the common difference.

### Step 2: Set up equations
From the above expressions, we can write:
\[ \frac{1}{a} - \frac{1}{b} = (p-q)D \]
\[ \frac{1}{b} - \frac{1}{c} = (q-r)D \]

### Step 3: Rearranging the equations
Rearranging the first equation gives:
\[ \frac{b-a}{ab} = (p-q)D \]
Rearranging the second equation gives:
\[ \frac{c-b}{bc} = (q-r)D \]

### Step 4: Equate the values of D
From both equations, we can express \( D \):
\[ D = \frac{b-a}{ab(p-q)} = \frac{c-b}{bc(q-r)} \]
Equating these two expressions for \( D \):
\[ \frac{b-a}{ab(p-q)} = \frac{c-b}{bc(q-r)} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ (b-a)bc(q-r) = (c-b)ab(p-q) \]
Expanding both sides leads to:
\[ bc(q-r)(b-a) = ab(c-b)(p-q) \]

### Step 6: Finding the determinant
Now we compute the determinant:
\[ \Delta = \begin{vmatrix} bc & ca & ab \\ p & q & r \\ 1 & 1 & 1 \end{vmatrix} \]
Using properties of determinants, we can perform column operations to simplify it. For example, we can replace the second column with the difference of the second and third columns, and the first column with the difference of the first and second columns.

After performing these operations, we can find that the determinant simplifies down to:
\[ \Delta = 0 \]

### Final Answer
Thus, the value of the determinant \( \Delta \) is:
\[ \Delta = 0 \]