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

Options

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

Answer

To solve the given problem, we need to evaluate the determinant on the left-hand side (LHS) and compare it with the right-hand side (RHS) to find the value of \( k \).

### Step-by-Step Solution:

1. **Write the Determinant**:
We need to evaluate 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 using the identity \((x+y)^2 = x^2 + 2xy + y^2\) and \((x-y)^2 = x^2 - 2xy + y^2\):
- Second row:
\[ (a+\lambda)^2 = a^2 + 2a\lambda + \lambda^2, \quad (b+\lambda)^2 = b^2 + 2b\lambda + \lambda^2, \quad (c+\lambda)^2 = c^2 + 2c\lambda + \lambda^2 \]
- Third row:
\[ (a-\lambda)^2 = a^2 - 2a\lambda + \lambda^2, \quad (b-\lambda)^2 = b^2 - 2b\lambda + \lambda^2, \quad (c-\lambda)^2 = c^2 - 2c\lambda + \lambda^2 \]

3. **Substituting Back into the Determinant**:
Substitute the expanded forms back into the determinant:
\[ D = \begin{vmatrix} a^2 & b^2 & c^2 \\ a^2 + 2a\lambda + \lambda^2 & b^2 + 2b\lambda + \lambda^2 & c^2 + 2c\lambda + \lambda^2 \\ a^2 - 2a\lambda + \lambda^2 & b^2 - 2b\lambda + \lambda^2 & c^2 - 2c\lambda + \lambda^2 \end{vmatrix} \]

4. **Row Operations**:
Perform row operations to simplify the determinant:
- Let \( R_2 \) be replaced by \( R_2 - R_1 \) and \( R_3 \) be replaced by \( R_3 - R_1 \):
\[ D = \begin{vmatrix} a^2 & b^2 & c^2 \\ 2a\lambda + \lambda^2 & 2b\lambda + \lambda^2 & 2c\lambda + \lambda^2 \\ -2a\lambda + \lambda^2 & -2b\lambda + \lambda^2 & -2c\lambda + \lambda^2 \end{vmatrix} \]

5. **Factor Out Common Terms**:
Factor out \( \lambda^2 \) from rows 2 and 3:
\[ D = \lambda^2 \begin{vmatrix} a^2 & b^2 & c^2 \\ 2a + 1 & 2b + 1 & 2c + 1 \\ -2a + 1 & -2b + 1 & -2c + 1 \end{vmatrix} \]

6. **Simplify Further**:
Continue simplifying the determinant using row operations until you reach a form that can be easily evaluated.

7. **Final Form**:
After simplification, we can express \( D \) in terms of \( \lambda \) and the original determinant:
\[ D = 4\lambda^2 \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} \]

8. **Equate to RHS**:
Given that \( D = k\lambda \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} \), we can compare coefficients:
\[ 4\lambda^2 = k\lambda \]

9. **Solve for \( k \)**:
Dividing both sides by \( \lambda \) (since \( \lambda \neq 0 \)):
\[ k = 4\lambda \]

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