If `a ,b ,c ,d ,e ,a n df` are in G.P. then the value of `|a^2d^2x b^2e^2y c^2f^2z|` depends on `xa n dy` b. `xa n dz` c. `ya n dz` d. independent of `x ,y ,a n dz`

Options

A. x and y
B. x and z
C. y and z
D. independent of x,y and z

Explanation

To solve the problem, we need to analyze the determinant given the conditions that \(a, b, c, d, e, f\) are in a geometric progression (G.P.).

### Step-by-step Solution:

1. **Understanding the G.P. Condition**:
Since \(a, b, c, d, e, f\) are in G.P., we can express them in terms of a common ratio \(r\):
- \(b = ar\)
- \(c = ar^2\)
- \(d = ar^3\)
- \(e = ar^4\)
- \(f = ar^5\)

2. **Setting up the Determinant**:
We need to evaluate the determinant:
\[ \begin{vmatrix} a^2 & b^2 & c^2 \\ d^2 & e^2 & f^2 \\ x & y & z \end{vmatrix} \]
Substituting the values of \(b, c, d, e, f\) in terms of \(a\) and \(r\):
\[ \begin{vmatrix} a^2 & (ar)^2 & (ar^2)^2 \\ (ar^3)^2 & (ar^4)^2 & (ar^5)^2 \\ x & y & z \end{vmatrix} = \begin{vmatrix} a^2 & a^2r^2 & a^2r^4 \\ a^2r^6 & a^2r^8 & a^2r^{10} \\ x & y & z \end{vmatrix} \]

3. **Factoring Out Common Terms**:
We can factor \(a^2\) from the first two rows:
\[ a^2 \cdot a^2 \cdot \begin{vmatrix} 1 & r^2 & r^4 \\ r^6 & r^8 & r^{10} \\ x & y & z \end{vmatrix} = a^4 \cdot \begin{vmatrix} 1 & r^2 & r^4 \\ r^6 & r^8 & r^{10} \\ x & y & z \end{vmatrix} \]

4. **Evaluating the Determinant**:
Now, we need to evaluate the determinant:
\[ \begin{vmatrix} 1 & r^2 & r^4 \\ r^6 & r^8 & r^{10} \\ x & y & z \end{vmatrix} \]
We can perform row operations or use properties of determinants to simplify this. Notice that the first two rows are linearly dependent (the second row is a multiple of the first), which implies that the determinant is equal to zero:
\[ = 0 \]

5. **Conclusion**:
Since the determinant evaluates to zero, it indicates that the rows are linearly dependent, meaning that the value of the determinant does not depend on \(x\), \(y\), or \(z\). Therefore, the correct answer is:
**d. independent of \(x\), \(y\), and \(z\)**.