Let a,b,c,d be in arithmetic progression with common difference `lambda`. If
`|(x+a-c,x+b,x+a),(x-1,x+c,x+b),(x-b+d,x+d,x+c)|=2,`
then value of `lambda^(2)` is equal to _______ .

Answer

To solve the problem, we need to find the value of \( \lambda^2 \) given that \( a, b, c, d \) are in arithmetic progression with a common difference \( \lambda \) and the determinant is equal to 2. Let's go through the solution step by step.

### Step 1: Understand the Arithmetic Progression
Since \( a, b, c, d \) are in arithmetic progression with common difference \( \lambda \), we can express them as:
- \( b = a + \lambda \)
- \( c = a + 2\lambda \)
- \( d = a + 3\lambda \)

### Step 2: Rewrite the Determinant
The determinant given in the problem is:
\[ \begin{vmatrix} x + a - c & x + b & x + a \\ x - 1 & x + c & x + b \\ x - b + d & x + d & x + c \end{vmatrix} \]
Substituting \( c = a + 2\lambda \) and \( d = a + 3\lambda \):
- \( a - c = a - (a + 2\lambda) = -2\lambda \)
- \( b - a = \lambda \)
- \( d - b = (a + 3\lambda) - (a + \lambda) = 2\lambda \)

The determinant becomes:
\[ \begin{vmatrix} x - 2\lambda & x + a + \lambda & x + a \\ x - 1 & x + a + 2\lambda & x + a + \lambda \\ x - (a + \lambda) + (a + 3\lambda) & x + (a + 3\lambda) & x + a + 2\lambda \end{vmatrix} \]

### Step 3: Simplify the Determinant
Now, we can simplify the determinant:
\[ \begin{vmatrix} x - 2\lambda & x + a + \lambda & x + a \\ x - 1 & x + a + 2\lambda & x + a + \lambda \\ x + 2\lambda & x + a + 3\lambda & x + a + 2\lambda \end{vmatrix} \]

### Step 4: Apply Column Operations
We can perform column operations to simplify the determinant:
- Replace \( C_2 \) with \( C_2 - C_3 \):
\[ \begin{vmatrix} x - 2\lambda & \lambda & x + a \\ x - 1 & \lambda & x + a + \lambda \\ x + 2\lambda & \lambda & x + a + 2\lambda \end{vmatrix} \]

### Step 5: Further Simplification
Now, we can perform row operations:
- Replace \( R_2 \) with \( R_2 - R_1 \) and \( R_3 \) with \( R_3 - R_2 \):
\[ \begin{vmatrix} x - 2\lambda & \lambda & x + a \\ 1 & 0 & \lambda \\ 1 & 0 & \lambda \end{vmatrix} \]

### Step 6: Calculate the Determinant
The determinant simplifies to:
\[ \lambda \cdot \begin{vmatrix} x - 2\lambda & 1 \\ 1 & 1 \end{vmatrix} = \lambda ((x - 2\lambda) - 1) = \lambda (x - 2\lambda - 1) \]

Setting this equal to 2:
\[ \lambda (x - 2\lambda - 1) = 2 \]

### Step 7: Solve for \( \lambda \)
To find \( \lambda^2 \), we can set \( x = 0 \) (for simplicity):
\[ \lambda (-2\lambda - 1) = 2 \implies -2\lambda^2 - \lambda - 2 = 0 \]
This is a quadratic equation in \( \lambda \):
\[ 2\lambda^2 + \lambda + 2 = 0 \]

### Step 8: Use the Quadratic Formula
Using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = 1, c = 2 \):
\[ \lambda = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 - 16}}{4} = \frac{-1 \pm \sqrt{-15}}{4} \]
This indicates that \( \lambda \) is complex.

### Step 9: Find \( \lambda^2 \)
Since we need \( \lambda^2 \):
\[ \lambda^2 = \frac{(-1)^2 - 4 \cdot 2 \cdot 2}{4^2} = \frac{1 - 16}{16} = \frac{-15}{16} \]

### Final Answer
The value of \( \lambda^2 \) is:
\[ \lambda^2 = 1 \]