If \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} = k\lambda \begin{vmatrix} a^2 & b^2 & c^2 \ a & b & c \ 1 & 1 & 1 \end{vmatrix}, \lambda \neq 0 then k is equal to :

Let the given determinant be denoted by Δ. Then
Δ = \begin{vmatrix} a^2 & b^2 & c^2 \ (a+λ)^2 & (b+λ)^2 & (c+λ)^2 \ (a-λ)^2 & (b-λ)^2 & (c-λ)^2 \end{vmatrix}
= \begin{vmatrix} a^2 & b^2 & c^2 \ a^2 + 2aλ + λ^2 & b^2 + 2bλ + λ^2 & c^2 + 2cλ + λ^2 \ a^2 - 2aλ + λ^2 & b^2 - 2bλ + λ^2 & c^2 - 2cλ + λ^2 \end{vmatrix}
Applying C_2 \to C_2 - C_1 and C_3 \to C_3 - C_1,
Δ = \begin{vmatrix} a^2 & b^2 - a^2 & c^2 - a^2 \ (a+λ)^2 & (b+λ)^2 - (a+λ)^2 & (c+λ)^2 - (a+λ)^2 \ (a-λ)^2 & (b-λ)^2 - (a-λ)^2 & (c-λ)^2 - (a-λ)^2 \end{vmatrix}
= \begin{vmatrix} a^2 & (b-a)(b+a) & (c-a)(c+a) \ (a+λ)^2 & (b-a)(b+a+2λ) & (c-a)(c+a+2λ) \ (a-λ)^2 & (b-a)(b+a-2λ) & (c-a)(c+a-2λ) \end{vmatrix}
Taking (b-a) and (c-a) common from C_2 and C_3 respectively,
Δ = (b-a)(c-a) \begin{vmatrix} a^2 & b+a & c+a \ (a+λ)^2 & b+a+2λ & c+a+2λ \ (a-λ)^2 & b+a-2λ & c+a-2λ \end{vmatrix}
Applying R_3 \to R_3 - R_1 and R_2 \to R_2 - R_1,
Δ = (b-a)(c-a) \begin{vmatrix} a^2 & b+a & c+a \ (a+λ)^2 - a^2 & 2λ & 2λ \ (a-λ)^2 - a^2 & -2λ & -2λ \end{vmatrix}
= (b-a)(c-a) \begin{vmatrix} a^2 & b+a & c+a \ 2aλ + λ^2 & 2λ & 2λ \ -2aλ + λ^2 & -2λ & -2λ \end{vmatrix}
Taking 2λ common from R_2 and R_3,
Δ = (b-a)(c-a)(2λ)^2 \begin{vmatrix} a^2 & b+a & c+a \ a+\frac{λ}{2} & 1 & 1 \ -a+\frac{λ}{2} & -1 & -1 \end{vmatrix}
= 4λ^2 (b-a)(c-a) \begin{vmatrix} a^2 & b+a & c+a \ a+\frac{λ}{2} & 1 & 1 \ -a+\frac{λ}{2} & -1 & -1 \end{vmatrix}
= 4λ^2 (b-a)(c-a) (2a)
= 8λ^2 a(b-a)(c-a)
If a=1, b=2, c=3, then
\begin{vmatrix} 1 & 4 & 9 \ 1 & 4 & 9 \ 1 & 4 & 9 \end{vmatrix} = 0
Given that Δ = kλ \begin{vmatrix} a^2 & b^2 & c^2 \ a & b & c \ 1 & 1 & 1 \end{vmatrix}
Comparing the two expressions, we can conclude that k = 4abc.