If x,y,z `in` R and `|{:(x,y^2,z^3),(x^4,y^5,z^6),(x^7,y^8,z^9):}|=2 `
then find the value of
`|{:(y^5z^6(z^3-y^3),x^4z^6(x^3-z^3),x^4y^5(y^3-x^3)),(y^2z^3(y^6-z^6),xz^3(z^6-x^6), xy^2(x^6-y^6)),(y^2z^3(z^3-y^3),xz^3(x^3-z^3),xy^2(y^3-x^3)):}|`

Answer

To solve the given problem, we need to find the value of the determinant:

\[ \Delta_2 = \left| \begin{array}{ccc} y^5 z^6 (z^3 - y^3) & x^4 z^6 (x^3 - z^3) & x^4 y^5 (y^3 - x^3) \\ y^2 z^3 (y^6 - z^6) & x z^3 (z^6 - x^6) & x y^2 (x^6 - y^6) \\ y^2 z^3 (z^3 - y^3) & x z^3 (x^3 - z^3) & x y^2 (y^3 - x^3) \end{array} \right| \]

### Step-by-step Solution:

1. **Understanding the Given Determinant**:
We know that the determinant of the first matrix is given as \( \Delta_1 = 2 \). The first matrix is:
\[ \Delta_1 = \left| \begin{array}{ccc} x & y^2 & z^3 \\ x^4 & y^5 & z^6 \\ x^7 & y^8 & z^9 \end{array} \right| = 2 \]

2. **Identifying the Structure of \(\Delta_2\)**:
The second determinant \(\Delta_2\) can be expressed in terms of the cofactors of \(\Delta_1\). The elements of \(\Delta_2\) are constructed from the differences of powers of \(x\), \(y\), and \(z\).

3. **Using Properties of Determinants**:
We can use the property of determinants that relates the determinant of a matrix to the determinant of its adjoint. For a square matrix \(A\) of order \(n\):
\[ \text{det}(\text{adj}(A)) = \text{det}(A)^{n-1} \]
Here, \(n = 3\) since we have a \(3 \times 3\) matrix.

4. **Finding the Value of \(\Delta_2\)**:
Since \(\Delta_2\) is related to the adjoint of the matrix corresponding to \(\Delta_1\), we can write:
\[ \Delta_2 = \text{det}(\text{adj}(A)) = \text{det}(A)^{3-1} = \text{det}(A)^2 \]
Given that \(\Delta_1 = 2\), we can substitute this value:
\[ \Delta_2 = (2)^2 = 4 \]

### Final Answer:
\[ \Delta_2 = 4 \]