Given, `2x - y + 2z = 2, x - 2y + z = -4, x + y+ lamda z = 4`,then the value of `lambda` such that the given system of equations has no solution, is

Options

A. 3
B. 1
C. 0
D. `-3`

Explanation

To find the value of \( \lambda \) such that the system of equations has no solution, we will analyze the given equations and use determinants.

### Given Equations:
1. \( 2x - y + 2z = 2 \)
2. \( x - 2y + z = -4 \)
3. \( x + y + \lambda z = 4 \)

### Step 1: Form the Coefficient Matrix
The coefficient matrix \( A \) for the system of equations is:

\[ A = \begin{bmatrix} 2 & -1 & 2 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{bmatrix} \]

### Step 2: Calculate the Determinant
We need to find the determinant \( D \) of the matrix \( A \). The determinant can be calculated using the formula:

\[ D = \begin{vmatrix} 2 & -1 & 2 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{vmatrix} \]

Using the determinant expansion along the first row:

\[ D = 2 \begin{vmatrix} -2 & 1 \\ 1 & \lambda \end{vmatrix} - (-1) \begin{vmatrix} 1 & 1 \\ 1 & \lambda \end{vmatrix} + 2 \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} -2 & 1 \\ 1 & \lambda \end{vmatrix} = (-2)(\lambda) - (1)(1) = -2\lambda - 1 \)

2. \( \begin{vmatrix} 1 & 1 \\ 1 & \lambda \end{vmatrix} = (1)(\lambda) - (1)(1) = \lambda - 1 \)

3. \( \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} = (1)(1) - (-2)(1) = 1 + 2 = 3 \)

Substituting back into the determinant equation:

\[ D = 2(-2\lambda - 1) + (\lambda - 1) + 2(3) \]

Expanding this gives:

\[ D = -4\lambda - 2 + \lambda - 1 + 6 \]

Combining like terms:

\[ D = -4\lambda + \lambda + 3 = -3\lambda + 3 \]

### Step 3: Set the Determinant to Zero
For the system of equations to have no solution, the determinant must be zero:

\[ -3\lambda + 3 = 0 \]

### Step 4: Solve for \( \lambda \)
Solving the equation:

\[ -3\lambda = -3 \implies \lambda = 1 \]

### Conclusion
The value of \( \lambda \) such that the given system of equations has no solution is:

\[ \lambda = 1 \]