The determinant `|(xp+y, x, y),(yp+z, y , z),(0, xp+y , yp+z)|=0` , if

Options

A. x,y,z are in AP
B. x , y , z are in GP
C. x,y,z are in HP
D. xy,yz,zx are in AP

Answer

To solve the determinant problem given by

\[ \begin{vmatrix} xp + y & x & y \\ yp + z & y & z \\ 0 & xp + y & yp + z \end{vmatrix} = 0, \]

we can follow these steps:

### Step 1: Write the Determinant
We start with the determinant:

\[ D = \begin{vmatrix} xp + y & x & y \\ yp + z & y & z \\ 0 & xp + y & yp + z \end{vmatrix} \]

### Step 2: Apply Row Operations
We can simplify the determinant using row operations. Let's perform the operation \( R_3 \leftarrow R_3 - R_2 \):

\[ D = \begin{vmatrix} xp + y & x & y \\ yp + z & y & z \\ 0 - (yp + z) & (xp + y - y) & (yp + z - z) \end{vmatrix} \]
This simplifies to:

\[ D = \begin{vmatrix} xp + y & x & y \\ yp + z & y & z \\ -(yp + z) & xp & yp \end{vmatrix} \]

### Step 3: Expand the Determinant
Now we can expand the determinant along the third row:

\[ D = -(-(yp + z)) \begin{vmatrix} x & y \\ y & z \end{vmatrix} + 0 + 0 \]

Calculating the 2x2 determinant:

\[ \begin{vmatrix} x & y \\ y & z \end{vmatrix} = xz - y^2 \]

Thus, we have:

\[ D = (yp + z)(xz - y^2) \]

### Step 4: Set the Determinant to Zero
For the determinant to be zero, we set:

\[ (yp + z)(xz - y^2) = 0 \]

This gives us two cases to consider:

1. \( yp + z = 0 \)
2. \( xz - y^2 = 0 \)

### Step 5: Solve the Second Case
From the second case, we have:

\[ xz = y^2 \]

This can be rearranged as:

\[ \frac{x}{y} = \frac{y}{z} \]

This indicates that \( x, y, z \) are in geometric progression (GP).

### Conclusion
Thus, the determinant \( |(xp+y, x, y),(yp+z, y , z),(0, xp+y , yp+z)|=0 \) if \( x, y, z \) are in GP.