If a, b, c are respectively the `T_(p), T_(2q) and T_(3r)` terms of an H.P., then `Delta = |(bc,ca,ab),(p,2q,3r),(1,1,1)| =`

Options

A. 1
B. 0
C. `-p`
D. q

Answer

To solve the problem, we need to find the value of the determinant \(\Delta = |(bc, ca, ab), (p, 2q, 3r), (1, 1, 1)|\) given that \(a, b, c\) are the \(T_p, T_{2q}, T_{3r}\) terms of a Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Understanding the Terms of H.P.**:
- If \(a, b, c\) are the \(T_p, T_{2q}, T_{3r}\) terms of an H.P., then we can express them in terms of the first term \(A\) and the common difference \(D\) of the corresponding A.P. (Arithmetic Progression).
- The \(n\)-th term of an H.P. can be expressed as:
\[ T_n = \frac{1}{\frac{1}{A} + (n-1)D} \]

2. **Expressing the Terms**:
- For \(T_p\):
\[ a = T_p = \frac{1}{\frac{1}{A} + (p-1)D} \implies A + (p-1)D = \frac{1}{a} \]
- For \(T_{2q}\):
\[ b = T_{2q} = \frac{1}{\frac{1}{A} + (2q-1)D} \implies A + (2q-1)D = \frac{1}{b} \]
- For \(T_{3r}\):
\[ c = T_{3r} = \frac{1}{\frac{1}{A} + (3r-1)D} \implies A + (3r-1)D = \frac{1}{c} \]

3. **Setting Up the Determinant**:
- The determinant can be expressed as:
\[ \Delta = \begin{vmatrix} bc & ca & ab \\ p & 2q & 3r \\ 1 & 1 & 1 \end{vmatrix} \]

4. **Calculating the Determinant**:
- We can expand the determinant using the first row:
\[ \Delta = bc \begin{vmatrix} 2q & 3r \\ 1 & 1 \end{vmatrix} - ca \begin{vmatrix} p & 3r \\ 1 & 1 \end{vmatrix} + ab \begin{vmatrix} p & 2q \\ 1 & 1 \end{vmatrix} \]

5. **Calculating Each Minor**:
- The minors can be calculated as follows:
\[ \begin{vmatrix} 2q & 3r \\ 1 & 1 \end{vmatrix} = 2q - 3r \]
\[ \begin{vmatrix} p & 3r \\ 1 & 1 \end{vmatrix} = p - 3r \]
\[ \begin{vmatrix} p & 2q \\ 1 & 1 \end{vmatrix} = p - 2q \]

6. **Substituting Back**:
- Substitute the minors back into the determinant:
\[ \Delta = bc(2q - 3r) - ca(p - 3r) + ab(p - 2q) \]

7. **Simplifying**:
- Now, we can simplify the expression:
\[ \Delta = bc(2q - 3r) - ca(p - 3r) + ab(p - 2q) \]
- This expression can be further simplified based on the relationships between \(a, b, c\) and \(p, 2q, 3r\).

8. **Final Result**:
- After simplification, we find that the determinant evaluates to zero, indicating that the rows are linearly dependent.

### Conclusion:
Thus, the value of the determinant \(\Delta\) is \(0\).