if `Delta= |{:(3,,4,,5,,x),(4,,5,,6,,y),(5,,6,,7,,z),(x,,y,,z,,0):}|=0` then

Options

A. `x,y,z " are in A.P."`
B. `x,y,z" are in G.P "`
C. `x, y, z " are in H.P "`
D. none of these

Explanation

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

\[ \Delta = \begin{vmatrix} 3 & 4 & 5 & x \\ 4 & 5 & 6 & y \\ 5 & 6 & 7 & z \\ x & y & z & 0 \end{vmatrix} \]

and set it equal to zero.

### Step 1: Apply Row Operations
We will perform row operations to simplify the determinant. We can use the operation \( R_1 \leftarrow R_1 + R_3 - 2R_2 \).

Calculating the new first row:
- First element: \( 3 + 5 - 2 \cdot 4 = 3 + 5 - 8 = 0 \)
- Second element: \( 4 + 6 - 2 \cdot 5 = 4 + 6 - 10 = 0 \)
- Third element: \( 5 + 7 - 2 \cdot 6 = 5 + 7 - 12 = 0 \)
- Fourth element: \( x + z - 2y \)

Thus, the determinant simplifies to:

\[ \Delta = \begin{vmatrix} 0 & 0 & 0 & x + z - 2y \\ 4 & 5 & 6 & y \\ 5 & 6 & 7 & z \\ x & y & z & 0 \end{vmatrix} \]

### Step 2: Expand the Determinant
Since the first row contains all zeros except for the last element, we can expand the determinant along the first row. The only term we get is:

\[ (x + z - 2y) \cdot \begin{vmatrix} 4 & 5 & 6 \\ 5 & 6 & 7 \\ y & z & 0 \end{vmatrix} \]

### Step 3: Calculate the 3x3 Determinant
Now we need to compute the determinant:

\[ \begin{vmatrix} 4 & 5 & 6 \\ 5 & 6 & 7 \\ y & z & 0 \end{vmatrix} \]

Using the formula for the determinant of a 3x3 matrix, we find:

\[ = 4 \begin{vmatrix} 6 & 7 \\ z & 0 \end{vmatrix} - 5 \begin{vmatrix} 5 & 7 \\ y & 0 \end{vmatrix} + 6 \begin{vmatrix} 5 & 6 \\ y & z \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 6 & 7 \\ z & 0 \end{vmatrix} = 6 \cdot 0 - 7 \cdot z = -7z \)

2. \( \begin{vmatrix} 5 & 7 \\ y & 0 \end{vmatrix} = 5 \cdot 0 - 7 \cdot y = -7y \)

3. \( \begin{vmatrix} 5 & 6 \\ y & z \end{vmatrix} = 5z - 6y \)

Putting it all together:

\[ \Delta = (x + z - 2y) \left( 4(-7z) - 5(-7y) + 6(5z - 6y) \right) \]

\[ = (x + z - 2y) \left( -28z + 35y + 30z - 36y \right) \]

\[ = (x + z - 2y) \left( 2z - y \right) \]

### Step 4: Set the Determinant Equal to Zero
Since we have \( \Delta = 0 \), we have two cases:

1. \( x + z - 2y = 0 \)
2. \( 2z - y = 0 \)

From the first equation, we can express \( x + z = 2y \), which implies:

\[ x + z = 2y \implies \frac{x + z}{2} = y \]

This indicates that \( x, y, z \) are in Arithmetic Progression (AP).

### Final Answer
Thus, the relation between \( x, y, z \) is:

\[ x + z = 2y \quad \text{(or equivalently, } y = \frac{x + z}{2}\text{)} \]