Let `Delta_(1)=|{:(ap^(2),2ap,1),(aq^(2),2aq,1),(ar^(2),2ar,1):}|` and `Delta_(2)=|{:(apq,a(p+q),1),(aqr,a(q+r),1),(arp,a(r+p),1):}|` then

Options

A. `Delta_(1)=Delta_(2)`
B. `Delta_(2)=2Delta_(1)`
C. `Delta_(1)=2Delta_(2)`
D. `Delta_(1)+2Delta_(2)=0`

Explanation

To solve the given problem, we need to evaluate the determinants \( \Delta_1 \) and \( \Delta_2 \) and then compare them based on the options provided.

### Step-by-Step Solution:

1. **Evaluate \( \Delta_1 \)**:
\[ \Delta_1 = \begin{vmatrix} ap^2 & 2ap & 1 \\ aq^2 & 2aq & 1 \\ ar^2 & 2ar & 1 \end{vmatrix} \]

We can factor out \( a \) from the first column and \( 2 \) from the second column:
\[ \Delta_1 = 2a \begin{vmatrix} p^2 & 1 \\ q^2 & 1 \\ r^2 & 1 \end{vmatrix} \]

2. **Row Operations**:
Now we apply row operations to simplify the determinant:
\[ R_1 \rightarrow R_1 - R_2 \quad \text{and} \quad R_2 \rightarrow R_2 - R_3 \]
This gives:
\[ \Delta_1 = 2a \begin{vmatrix} p^2 - q^2 & 0 \\ q^2 - r^2 & 0 \\ r^2 & 1 \end{vmatrix} \]

3. **Factor the Determinant**:
The determinant can now be factored as:
\[ \Delta_1 = 2a \cdot (p - q)(p + q) \cdot (q - r)(q + r) \cdot (r - p)(r + p) \]

4. **Evaluate \( \Delta_2 \)**:
\[ \Delta_2 = \begin{vmatrix} apq & a(p + q) & 1 \\ aqr & a(q + r) & 1 \\ arp & a(r + p) & 1 \end{vmatrix} \]

Again, we can factor out \( a \) from the first column:
\[ \Delta_2 = a^2 \begin{vmatrix} pq & p + q & 1 \\ qr & q + r & 1 \\ rp & r + p & 1 \end{vmatrix} \]

5. **Row Operations**:
Applying the same row operations:
\[ R_1 \rightarrow R_1 - R_2 \quad \text{and} \quad R_2 \rightarrow R_2 - R_3 \]
This gives:
\[ \Delta_2 = a^2 \begin{vmatrix} pq - qr & p + q - (q + r) & 0 \\ qr - rp & q + r - (r + p) & 0 \\ rp & r + p & 1 \end{vmatrix} \]

6. **Factor the Determinant**:
The determinant simplifies to:
\[ \Delta_2 = a^2 \cdot (p - r)(q - p)(q - r) \]

7. **Compare \( \Delta_1 \) and \( \Delta_2 \)**:
From the calculations, we have:
\[ \Delta_1 = 2a^2 (p - q)(q - r)(r - p) \]
\[ \Delta_2 = a^2 (p - r)(q - p)(q - r) \]

Therefore, we can conclude:
\[ \Delta_1 = 2 \Delta_2 \]

### Final Conclusion:
The correct option is that \( \Delta_1 = 2 \Delta_2 \).