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

Answer

To solve the given determinant equation, we will follow these steps:

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

### Step 2: Expand the Second and Third Rows
We expand the second and third rows:
\[ (a + \lambda)^2 = a^2 + 2a\lambda + \lambda^2 \]
\[ (b + \lambda)^2 = b^2 + 2b\lambda + \lambda^2 \]
\[ (c + \lambda)^2 = c^2 + 2c\lambda + \lambda^2 \]
Similarly,
\[ (a - \lambda)^2 = a^2 - 2a\lambda + \lambda^2 \]
\[ (b - \lambda)^2 = b^2 - 2b\lambda + \lambda^2 \]
\[ (c - \lambda)^2 = c^2 - 2c\lambda + \lambda^2 \]

### Step 3: Substitute Back into the Determinant
Now substituting these expansions back into the determinant:
\[ D = \begin{vmatrix} a^2 & b^2 & c^2 \\ a^2 + 2a\lambda + \lambda^2 & b^2 + 2b\lambda + \lambda^2 & c^2 + 2c\lambda + \lambda^2 \\ a^2 - 2a\lambda + \lambda^2 & b^2 - 2b\lambda + \lambda^2 & c^2 - 2c\lambda + \lambda^2 \end{vmatrix} \]

### Step 4: Row Operations
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:
\[ D = \begin{vmatrix} a^2 & b^2 & c^2 \\ 2a\lambda + \lambda^2 & 2b\lambda + \lambda^2 & 2c\lambda + \lambda^2 \\ -2a\lambda & -2b\lambda & -2c\lambda \end{vmatrix} \]

### Step 5: Factor Out Common Terms
Notice that we can factor out \(\lambda\) from the second and third rows:
\[ D = \lambda \begin{vmatrix} a^2 & b^2 & c^2 \\ 2a + \lambda & 2b + \lambda & 2c + \lambda \\ -2a & -2b & -2c \end{vmatrix} \]

### Step 6: Evaluate the Determinant
Now we can evaluate the determinant. The determinant can be simplified further by using properties of determinants. The final result will yield:
\[ D = k \lambda |(a^2, b^2, c^2), (a, b, c), (1, 1, 1)| \]

### Step 7: Find the Value of k
From the evaluation, we find that \(k\) is equal to 4. Thus:
\[ k = 4 \]

### Final Answer
The value of \(k\) is:
\[ \boxed{4} \]