If `Delta=|(a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)| and A_2,B_2,C_2` are respectively cofactors of `a_2,b_2,c_2` then `a_1A_2+b_1B_2+c_1C_2` is

Options

A. `a_1A_1 + b_1B_1 + c_1C_1=Delta`
B. `a_2A_2 + b_2B_2+ c_2C_2=Delta`
C. `a_3A_3+b_3B_3+c_3C_3=Delta`
D. `a_1A_2+b_1B_2 + c_1 C_2=0`

Answer

To solve the problem, we need to evaluate the expression \( a_1 A_2 + b_1 B_2 + c_1 C_2 \) given that \( \Delta = |(a_1, b_1, c_1), (a_2, b_2, c_2), (a_3, b_3, c_3)| \) and \( A_2, B_2, C_2 \) are the cofactors of \( a_2, b_2, c_2 \) respectively.

### Step-by-Step Solution:

1. **Understanding the Determinant**:
\[ \Delta = |(a_1, b_1, c_1), (a_2, b_2, c_2), (a_3, b_3, c_3)| \]
This determinant represents a 3x3 matrix formed by the vectors \( (a_1, b_1, c_1) \), \( (a_2, b_2, c_2) \), and \( (a_3, b_3, c_3) \).

**Hint**: Recall that the determinant can be interpreted as a volume of the parallelepiped formed by the vectors.

2. **Cofactors**:
The cofactors \( A_2, B_2, C_2 \) are defined as the determinants of the 2x2 matrices obtained by removing the row and column of the respective element \( a_2, b_2, c_2 \) from the original matrix.

**Hint**: Remember that the cofactor \( A_2 \) is calculated as \( (-1)^{i+j} \times \) (determinant of the submatrix), where \( i \) and \( j \) are the row and column indices of the element.

3. **Using the Property of Determinants**:
There is a property of determinants that states that the sum of the products of the elements of any row with the cofactors of the corresponding elements of another row equals zero. Here, we can apply this property.

**Hint**: This property can be expressed mathematically as:
\[ a_1 A_2 + b_1 B_2 + c_1 C_2 = 0 \]

4. **Conclusion**:
From the property mentioned above, we conclude that:
\[ a_1 A_2 + b_1 B_2 + c_1 C_2 = 0 \]

Therefore, the final answer is:
\[ a_1 A_2 + b_1 B_2 + c_1 C_2 = 0 \]

### Final Answer:
\[ \boxed{0} \]