If x, y , z are in A.P., then the value of the det (A) is , where
`A = [(4,5,6,x),(5,6,7,y),(6,7,8,z),(x,y,z,0)]`

Options

A. 0
B. 1
C. 2
D. none of these

Answer

To find the value of the determinant of the matrix \( A \) given that \( x, y, z \) are in arithmetic progression (A.P.), we can follow these steps:

### Step 1: Understand the condition of A.P.
Since \( x, y, z \) are in A.P., we can express \( y \) in terms of \( x \) and \( z \):
\[ y = \frac{x + z}{2} \]

### Step 2: Write the matrix \( A \)
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 4 & 5 & 6 & x \\ 5 & 6 & 7 & y \\ 6 & 7 & 8 & z \\ x & y & z & 0 \end{pmatrix} \]

### Step 3: Substitute \( y \) in the matrix
Using the expression for \( y \):
\[ A = \begin{pmatrix} 4 & 5 & 6 & x \\ 5 & 6 & 7 & \frac{x + z}{2} \\ 6 & 7 & 8 & z \\ x & \frac{x + z}{2} & z & 0 \end{pmatrix} \]

### Step 4: Factor out constants from the determinant
We can factor out \( \frac{1}{2} \) from the fourth column:
\[ \text{det}(A) = \frac{1}{2} \text{det}\begin{pmatrix} 4 & 5 & 6 & 2x \\ 5 & 6 & 7 & x + z \\ 6 & 7 & 8 & 2z \\ 2x & x + z & 2z & 0 \end{pmatrix} \]

### Step 5: Break the determinant into two parts
We can express the determinant as the sum of two determinants:
\[ \text{det}(A) = \frac{1}{2} \left( \text{det}\begin{pmatrix} 4 & 5 & 6 & 2x \\ 5 & 6 & 7 & x + z \\ 6 & 7 & 8 & 2z \\ 2x & x + z & 2z & 0 \end{pmatrix} + \text{det}\begin{pmatrix} 4 & 5 & 6 & x \\ 5 & 6 & 7 & y \\ 6 & 7 & 8 & z \\ x & y & z & 0 \end{pmatrix} \right) \]

### Step 6: Identify identical columns
Notice that the last two columns of the determinant are identical:
\[ \begin{pmatrix} 2x & x + z \\ 2z & z \\ 0 \end{pmatrix} \]
Since two columns of the determinant are identical, the value of the determinant is zero.

### Step 7: Conclude the value of the determinant
Thus, we conclude:
\[ \text{det}(A) = 0 \]

### Final Answer
The value of the determinant \( A \) is:
\[ \boxed{0} \]