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 given determinant problem, we need to analyze the determinant:

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

We are given that this determinant equals zero, i.e., \(\Delta = 0\).

### Step 1: Apply Row Transformation
We will perform a row operation to simplify the determinant. We can replace the first row \(R_1\) with \(R_1 - 2R_2 + R_3\):

\[ R_1 \rightarrow R_1 - 2R_2 + R_3 \]

This gives us:

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

### Step 2: Expand the Determinant
Now, we can expand the determinant along the first row. Since the first element of the first row is zero, the expansion simplifies to:

\[ \Delta = 0 + (-1)^{1+2} \cdot 4 \cdot \begin{vmatrix} 4 & 6 & y \\ 5 & 7 & z \\ y & z & 0 \end{vmatrix} + (-1)^{1+3} \cdot 5 \cdot \begin{vmatrix} 4 & 5 & y \\ 5 & 6 & z \\ x & y & 0 \end{vmatrix} \]

### Step 3: Calculate the Remaining Determinants
We need to calculate the two remaining \(3 \times 3\) determinants.

1. For the first determinant:

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

Using the determinant formula, we can compute this determinant.

2. For the second determinant:

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

Similarly, we compute this determinant.

### Step 4: Set the Determinant to Zero
After calculating both determinants, we will set the entire expression for \(\Delta\) to zero:

\[ \Delta = - (x - 2y + z) \cdot \text{(some expression)} = 0 \]

This leads us to the condition:

\[ x - 2y + z = 0 \]

### Step 5: Analyze the Condition
The equation \(x - 2y + z = 0\) can be rearranged to:

\[ x + z = 2y \]

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

### Conclusion
Thus, we conclude that \(x\), \(y\), and \(z\) are in AP.