If `A_1B_1C_1,A_2B_2C_2` and `A_3B_3C_3` are three three-digit numbers each of which is divisible by `lamda` then `Delta = |(A_1,B_1,C_1),(A_2,B_2,C_2),(A_3,B_3,C_3)|` is divisible by

Options

A. `lamda`
B. `lamda^2`
C. `2lamda`
D. none

Answer

To solve the problem, we need to show that the determinant \( \Delta = |(A_1, B_1, C_1), (A_2, B_2, C_2), (A_3, B_3, C_3)| \) is divisible by \( \lambda \) given that each of the three-digit numbers \( A_1B_1C_1, A_2B_2C_2, A_3B_3C_3 \) is divisible by \( \lambda \).

### Step-by-Step Solution:

1. **Express the three-digit numbers in terms of \( \lambda \)**:
Each three-digit number can be expressed as:
\[ A_1B_1C_1 = 100A_1 + 10B_1 + C_1 = m_1 \lambda \]
\[ A_2B_2C_2 = 100A_2 + 10B_2 + C_2 = m_2 \lambda \]
\[ A_3B_3C_3 = 100A_3 + 10B_3 + C_3 = m_3 \lambda \]
where \( m_1, m_2, m_3 \) are integers.

**Hint**: Recognize that each three-digit number can be represented as a linear combination of its digits multiplied by powers of 10.

2. **Set up the determinant**:
The determinant \( \Delta \) can be written as:
\[ \Delta = \begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} \]

3. **Substitute the expressions for \( C_1, C_2, C_3 \)**:
We can express the last column in terms of \( C_1, C_2, C_3 \):
\[ C_1 = m_1 \lambda - 100A_1 - 10B_1 \]
\[ C_2 = m_2 \lambda - 100A_2 - 10B_2 \]
\[ C_3 = m_3 \lambda - 100A_3 - 10B_3 \]

4. **Factor out \( \lambda \)**:
The determinant can be expressed as:
\[ \Delta = \begin{vmatrix} A_1 & B_1 & m_1 \lambda - 100A_1 - 10B_1 \\ A_2 & B_2 & m_2 \lambda - 100A_2 - 10B_2 \\ A_3 & B_3 & m_3 \lambda - 100A_3 - 10B_3 \end{vmatrix} \]

By using properties of determinants, we can factor out \( \lambda \):
\[ \Delta = \lambda \begin{vmatrix} A_1 & B_1 & 1 \\ A_2 & B_2 & 1 \\ A_3 & B_3 & 1 \end{vmatrix} \]

5. **Conclusion**:
Since \( \Delta \) contains \( \lambda \) as a factor, it follows that \( \Delta \) is divisible by \( \lambda \).

Therefore, the final answer is that \( \Delta \) is divisible by \( \lambda \).

### Final Answer:
\[ \Delta \text{ is divisible by } \lambda. \]