If `t_1, t_2 ,t_3` are in A.P. (a,d) and `T_1,T_2,T_3` are in H.P. then the value of
`Delta=|(t_1-T_1,t_1-T_2,t_1-T_3),(t_2-T_1,t_2-T_2,t_2-T_3),(t_3-T_1,t_3-T_2,t_3-T_3)|`

Options

A. a + d
B. `a^2+d^2`
C. 0
D. none

Answer

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

\[ \Delta = \begin{vmatrix} t_1 - T_1 & t_1 - T_2 & t_1 - T_3 \\ t_2 - T_1 & t_2 - T_2 & t_2 - T_3 \\ t_3 - T_1 & t_3 - T_2 & t_3 - T_3 \end{vmatrix} \]

### Step 1: Understanding the terms
Given that \( t_1, t_2, t_3 \) are in arithmetic progression (A.P.), we can express them as:
- \( t_1 = a \)
- \( t_2 = a + d \)
- \( t_3 = a + 2d \)

Also, since \( T_1, T_2, T_3 \) are in harmonic progression (H.P.), we can express them as:
- \( T_1 = \frac{2ab}{a+b} \)
- \( T_2 = \frac{2bc}{b+c} \)
- \( T_3 = \frac{2ac}{a+c} \)

However, for the determinant, we will keep \( T_1, T_2, T_3 \) as they are for now.

### Step 2: Substitute the values into the determinant
Substituting \( t_1, t_2, t_3 \) into the determinant gives:

\[ \Delta = \begin{vmatrix} a - T_1 & a - T_2 & a - T_3 \\ (a + d) - T_1 & (a + d) - T_2 & (a + d) - T_3 \\ (a + 2d) - T_1 & (a + 2d) - T_2 & (a + 2d) - T_3 \end{vmatrix} \]

### Step 3: Row operations
We can perform row operations to simplify the determinant. We will replace \( R_1 \) with \( R_1 - R_3 \) and \( R_2 \) with \( R_2 - R_3 \):

\[ \Delta = \begin{vmatrix} (a - T_1) - ((a + 2d) - T_1) & (a - T_2) - ((a + 2d) - T_2) & (a - T_3) - ((a + 2d) - T_3) \\ (a + d) - T_1 & (a + d) - T_2 & (a + d) - T_3 \\ (a + 2d) - T_1 & (a + 2d) - T_2 & (a + 2d) - T_3 \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} -2d & -2d & -2d \\ (a + d) - T_1 & (a + d) - T_2 & (a + d) - T_3 \\ (a + 2d) - T_1 & (a + 2d) - T_2 & (a + 2d) - T_3 \end{vmatrix} \]

### Step 4: Factor out common terms
Now we can factor out \(-2d\) from the first row:

\[ \Delta = -2d \begin{vmatrix} 1 & 1 & 1 \\ (a + d) - T_1 & (a + d) - T_2 & (a + d) - T_3 \\ (a + 2d) - T_1 & (a + 2d) - T_2 & (a + 2d) - T_3 \end{vmatrix} \]

### Step 5: Evaluate the determinant
Notice that the first row is now all ones. We can perform another row operation by replacing \( R_2 \) with \( R_2 - R_3 \):

\[ \Delta = -2d \begin{vmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ (a + 2d) - T_1 & (a + 2d) - T_2 & (a + 2d) - T_3 \end{vmatrix} \]

Since the second row is now all zeros, the value of the determinant is:

\[ \Delta = 0 \]

### Final Answer
Thus, the value of \(\Delta\) is:

\[ \Delta = 0 \]