Suppose `a_(1),a_(2),a_(3)` are in A.P. and `b_(1),b_(2),b_(3)` are in H.P. and let `/_\=|(a_(1)-b_(1),a_(1)-b_(2),a_(1)-b_(3)),(a_(2)-b_(1),a_(2)-b_(2),a_(2)-b_(3)),(a_(3)-b_(1),a_(3)-b_(2),a_(3)-b_(3))|`
then

Options

A. `/_\ "is independent of "a_(1),a_(2),a_(3),b_(1),b_(2),b_(3)`
B. `a_(1)-/_\,a_(2)-2/_\,a_(3)-3/_\` are in H.P.
C. `b_(1)+/_\,b_(2)+/_\^(2),b^(3)+/_\` are in H.P.
D. none of these

Explanation

To solve the problem, we need to analyze the given conditions and the determinant defined by the differences between the terms in the arithmetic progression (A.P.) and harmonic progression (H.P.).

### Step-by-Step Solution:

1. **Understanding A.P. and H.P.**:
- Let \( a_1, a_2, a_3 \) be in A.P. This means:
\[ a_2 - a_1 = a_3 - a_2 = d \quad \text{(common difference)} \]
- Let \( b_1, b_2, b_3 \) be in H.P. This implies that:
\[ \frac{1}{b_1}, \frac{1}{b_2}, \frac{1}{b_3} \text{ are in A.P.} \]
Therefore,
\[ 2 \cdot \frac{1}{b_2} = \frac{1}{b_1} + \frac{1}{b_3} \implies b_1 + b_3 = 2b_2 \]

2. **Setting Up the Determinant**:
- The determinant \( \Delta \) is defined as:
\[ \Delta = \begin{vmatrix} a_1 - b_1 & a_1 - b_2 & a_1 - b_3 \\ a_2 - b_1 & a_2 - b_2 & a_2 - b_3 \\ a_3 - b_1 & a_3 - b_2 & a_3 - b_3 \end{vmatrix} \]

3. **Substituting Values**:
- Substitute \( a_2 = a_1 + d \) and \( a_3 = a_1 + 2d \) into the determinant:
\[ \Delta = \begin{vmatrix} a_1 - b_1 & a_1 - b_2 & a_1 - b_3 \\ (a_1 + d) - b_1 & (a_1 + d) - b_2 & (a_1 + d) - b_3 \\ (a_1 + 2d) - b_1 & (a_1 + 2d) - b_2 & (a_1 + 2d) - b_3 \end{vmatrix} \]
- Simplifying, we have:
\[ \Delta = \begin{vmatrix} a_1 - b_1 & a_1 - b_2 & a_1 - b_3 \\ d & d & d \\ 2d & 2d & 2d \end{vmatrix} \]

4. **Calculating the Determinant**:
- Notice that the second and third rows are proportional:
- Row 2 is \( d \) times Row 1.
- Row 3 is \( 2d \) times Row 1.
- Since two rows of the determinant are proportional, the determinant evaluates to zero:
\[ \Delta = 0 \]

5. **Conclusion**:
- Since \( \Delta = 0 \), it implies that the value of the determinant does not depend on the specific values of \( a_1, a_2, a_3, b_1, b_2, b_3 \).

### Final Result:
\[ \Delta < 0 \implies \text{The determinant is independent of } a_1, a_2, a_3, b_1, b_2, b_3. \]