Let `Delta_(1)=|{:(Ax,x^(2),1),(By,y^(2),1),(Cz,z^(2),1):}|` and `Delta_(2)=|{:(A,B,C),(x,y,z),(yz,zx,xy):}|`, then :

Options

A. `Delta_(1)=-Delta`
B. `Delta ne Delta_(1)`
C. `Delta-Delta_(1)=0`
D. None of these

Answer

To solve the problem, we need to analyze the two determinants given:

1. \( \Delta_1 = \begin{vmatrix} Ax & x^2 & 1 \\ By & y^2 & 1 \\ Cz & z^2 & 1 \end{vmatrix} \)
2. \( \Delta_2 = \begin{vmatrix} A & B & C \\ x & y & z \\ yz & zx & xy \end{vmatrix} \)

We will find the relationship between these two determinants step by step.

### Step 1: Analyze \( \Delta_2 \)

We start with the determinant \( \Delta_2 \):
\[ \Delta_2 = \begin{vmatrix} A & B & C \\ x & y & z \\ yz & zx & xy \end{vmatrix} \]

### Step 2: Transpose \( \Delta_2 \)

Using the property of determinants that states the determinant of a matrix is equal to the determinant of its transpose, we have:
\[ \Delta_2 = \begin{vmatrix} A & x & yz \\ B & y & zx \\ C & z & xy \end{vmatrix} \]

### Step 3: Row Operations on \( \Delta_2 \)

Next, we will perform row operations. We can multiply the first row by \( x \), the second row by \( y \), and the third row by \( z \):
\[ \Delta_2 = \begin{vmatrix} Ax & By & Cz \\ x & y & z \\ yz & zx & xy \end{vmatrix} \]

### Step 4: Factor Out \( xyz \)

Since we multiplied each row by \( x \), \( y \), and \( z \), we need to account for this by dividing by \( xyz \):
\[ \Delta_2 = \frac{1}{xyz} \begin{vmatrix} Ax & By & Cz \\ x & y & z \\ yz & zx & xy \end{vmatrix} \]

### Step 5: Simplify the Determinant

Now we can factor out \( xyz \) from the third column:
\[ \Delta_2 = \frac{1}{xyz} \begin{vmatrix} Ax & By & Cz \\ x & y & z \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 6: Compare with \( \Delta_1 \)

Now we can see that:
\[ \Delta_1 = \begin{vmatrix} Ax & x^2 & 1 \\ By & y^2 & 1 \\ Cz & z^2 & 1 \end{vmatrix} \]

### Step 7: Establish the Relationship

From the previous steps, we can conclude that:
\[ \Delta_2 = \Delta_1 \]

Thus, we can write:
\[ \Delta_2 - \Delta_1 = 0 \]

### Conclusion

The relationship between the two determinants is:
\[ \Delta_2 = \Delta_1 \]