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:

$$\Delta_1 = \begin{vmatrix} a^2 & b^2 & c^2 \\ (a + \lambda)^2 & (b + \lambda)^2 & (c + \lambda)^2 \\ (a - \lambda)^2 & (b - \lambda)^2 & (c - \lambda)^2 \end{vmatrix}$$

$$\Delta_1 = \begin{vmatrix} a^2 & b^2 & c^2 \\ a^2 + 2a\lambda + \lambda^2 & b^2 + 2b\lambda + \lambda^2 & c^2 + 2c\lambda + \lambda^2 \\ a^2 - 2a\lambda + \lambda^2 & b^2 - 2b\lambda + \lambda^2 & c^2 - 2c\lambda + \lambda^2 \end{vmatrix}$$

Operation 1: $$R_2 \to R_2 - R_3$$

The elements of $$R_2$$ become $$(a^2 + 2a\lambda + \lambda^2) - (a^2 - 2a\lambda + \lambda^2) = 4a\lambda$$.

$$\Delta_1 = \begin{vmatrix} a^2 & b^2 & c^2 \\ 4a\lambda & 4b\lambda & 4c\lambda \\ a^2 - 2a\lambda + \lambda^2 & b^2 - 2b\lambda + \lambda^2 & c^2 - 2c\lambda + \lambda^2 \end{vmatrix}$$

For the first column: $$(a^2 - 2a\lambda + \lambda^2) + \frac{1}{2}(4a\lambda) - a^2 = \lambda^2$$.

$$\Delta_1 = \begin{vmatrix} a^2 & b^2 & c^2 \\ 4a\lambda & 4b\lambda & 4c\lambda \\ \lambda^2 & \lambda^2 & \lambda^2 \end{vmatrix}$$

$$\Delta_1 = (4\lambda)(\lambda^2) \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} = 4\lambda^3 \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix}$$

$$4\lambda^3 \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix} = k\lambda \begin{vmatrix} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{vmatrix}$$

$$k = 4\lambda^2$$