If
`|(a^(2),b^(2),c^(2)),((a+lamda)^(2),(b+lamda)^(2),(c+lamda)^(2)),((a-lamda)^(2),(b-lamda)^(2),(c-lamda)^(2))|=k lamda|(a^(2),b^(2),c^(2)),(a,b,c),(1,1,1)|`
`lamda!=0` then k is equal to

Options

A. `4 lamda abc`
B. `-4 lamda abc`
C. `4lamda^(2)`
D. `-4 lamda^(2)`

Answer

To solve the given determinant equation, we will follow a systematic approach to simplify the determinant and find the value of \( k \).

### Step-by-Step Solution:

1. **Write the Determinant**:
We start with the determinant:
\[ D = \begin{vmatrix} a^2 & b^2 & c^2 \\ (a + \lambda)^2 & (b + \lambda)^2 & (c + \lambda)^2 \\ (a - \lambda)^2 & (b - \lambda)^2 & (c - \lambda)^2 \end{vmatrix} \]

2. **Expand the Rows**:
Expand the second and third rows:
\[ (a + \lambda)^2 = a^2 + 2a\lambda + \lambda^2 \]
\[ (a - \lambda)^2 = a^2 - 2a\lambda + \lambda^2 \]
Thus, we rewrite the determinant:
\[ D = \begin{vmatrix} a^2 & b^2 & c^2 \\ a^2 + 2a\lambda + \lambda^2 & b^2 + 2b\lambda + \lambda^2 & c^2 + 2c\lambda + \lambda^2 \\ a^2 - 2a\lambda + \lambda^2 & b^2 - 2b\lambda + \lambda^2 & c^2 - 2c\lambda + \lambda^2 \end{vmatrix} \]

3. **Row Operations**:
Now, we perform row operations to simplify the determinant. We can subtract the first row from the second and third rows:
\[ R_2 \rightarrow R_2 - R_1 \]
\[ R_3 \rightarrow R_3 - R_1 \]
This gives us:
\[ D = \begin{vmatrix} a^2 & b^2 & c^2 \\ 2a\lambda + \lambda^2 & 2b\lambda + \lambda^2 & 2c\lambda + \lambda^2 \\ -2a\lambda + \lambda^2 & -2b\lambda + \lambda^2 & -2c\lambda + \lambda^2 \end{vmatrix} \]

4. **Factor Out Common Terms**:
Notice that each term in the second and third rows can be factored:
\[ D = \begin{vmatrix} a^2 & b^2 & c^2 \\ \lambda^2 + 2a\lambda & \lambda^2 + 2b\lambda & \lambda^2 + 2c\lambda \\ \lambda^2 - 2a\lambda & \lambda^2 - 2b\lambda & \lambda^2 - 2c\lambda \end{vmatrix} \]

5. **Further Simplification**:
We can express the determinant in terms of \( \lambda \):
\[ D = \lambda^2 \begin{vmatrix} a^2 & b^2 & c^2 \\ 2a & 2b & 2c \\ -2a & -2b & -2c \end{vmatrix} \]

6. **Calculate the Determinant**:
The determinant simplifies to:
\[ D = 4\lambda^2 \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} \]

7. **Set Equal to Given Expression**:
According to the problem statement:
\[ D = k\lambda \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} \]
Thus, we have:
\[ 4\lambda^2 \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} = k\lambda \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} \]

8. **Divide Both Sides**:
Since \( \lambda \neq 0 \), we can divide both sides by \( \lambda \):
\[ 4\lambda \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} = k \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} \]

9. **Final Comparison**:
By comparing coefficients, we find:
\[ k = 4\lambda \]

### Conclusion:
Thus, the value of \( k \) is \( 4 \).