Let $$a, b, c, d$$ be in arithmetic progression with common difference $$\lambda$$. If
$$\begin{vmatrix} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{vmatrix} = 2$$,
then value of $$\lambda^2$$ is equal to ___.

Since $$a, b, c, d$$ are in AP with common difference $$\lambda$$, we write $$b = a+\lambda$$, $$c = a+2\lambda$$, $$d = a+3\lambda$$. Substituting into the determinant:

$$\Delta = \begin{vmatrix} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{vmatrix} = \begin{vmatrix} x-2\lambda & x+a+\lambda & x+a \\ x-1 & x+a+2\lambda & x+a+\lambda \\ x+2\lambda & x+a+3\lambda & x+a+2\lambda \end{vmatrix}$$

Apply $$R_1 \to R_1 - R_2$$ and $$R_2 \to R_2 - R_3$$:

$$R_1 - R_2 = (x-2\lambda-(x-1),\ (x+a+\lambda)-(x+a+2\lambda),\ (x+a)-(x+a+\lambda)) = (1-2\lambda,\ -\lambda,\ -\lambda)$$

$$R_2 - R_3 = (x-1-(x+2\lambda),\ (x+a+2\lambda)-(x+a+3\lambda),\ (x+a+\lambda)-(x+a+2\lambda)) = (-1-2\lambda,\ -\lambda,\ -\lambda)$$

Now apply $$C_3 \to C_3 - C_2$$ to simplify the third column. In rows 1 and 2, $$C_3 - C_2 = -\lambda - (-\lambda) = 0$$. In row 3, $$C_3 - C_2 = (x+a+2\lambda) - (x+a+3\lambda) = -\lambda$$. The determinant becomes:

$$\Delta = \begin{vmatrix} 1-2\lambda & -\lambda & 0 \\ -1-2\lambda & -\lambda & 0 \\ x+2\lambda & x+a+3\lambda & -\lambda \end{vmatrix}$$

Expanding along the third column (only the $$(3,3)$$ entry is nonzero):

$$\Delta = (-\lambda) \cdot \begin{vmatrix} 1-2\lambda & -\lambda \\ -1-2\lambda & -\lambda \end{vmatrix} = (-\lambda)\left[(-\lambda)(1-2\lambda) - (-\lambda)(-1-2\lambda)\right]$$

$$= (-\lambda)\left[-\lambda + 2\lambda^2 - \lambda - 2\lambda^2\right] = (-\lambda)(-2\lambda) = 2\lambda^2$$

Setting $$\Delta = 2$$: $$2\lambda^2 = 2 \implies \lambda^2 = 1$$.