`lambda x + 2y + 2z = 5, 2lambda x + 3y + 5z = 8, 4x + lambda y + 6z = 10` for the system of equation check the correct option.

Options

A. no solution when `lamda=8`
B. no solution when `lamda=2`
C. a unique solution when `lamda=-8`
D. infinitley many solutions when `lamda=2`

Answer

To solve the system of equations given by:

1. \( \lambda x + 2y + 2z = 5 \) (Equation 1)
2. \( 2\lambda x + 3y + 5z = 8 \) (Equation 2)
3. \( 4x + \lambda y + 6z = 10 \) (Equation 3)

we will follow these steps:

### Step 1: Write the system in matrix form
The system can be represented in matrix form as:

\[ \begin{bmatrix} \lambda & 2 & 2 \\ 2\lambda & 3 & 5 \\ 4 & \lambda & 6 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 10 \end{bmatrix} \]

### Step 2: Calculate the determinant \( \Delta \)
The determinant \( \Delta \) of the coefficient matrix is given by:

\[ \Delta = \begin{vmatrix} \lambda & 2 & 2 \\ 2\lambda & 3 & 5 \\ 4 & \lambda & 6 \end{vmatrix} \]

Calculating this determinant, we expand it:

\[ \Delta = \lambda \begin{vmatrix} 3 & 5 \\ \lambda & 6 \end{vmatrix} - 2 \begin{vmatrix} 2\lambda & 5 \\ 4 & 6 \end{vmatrix} + 2 \begin{vmatrix} 2\lambda & 3 \\ 4 & \lambda \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 3 & 5 \\ \lambda & 6 \end{vmatrix} = 3 \cdot 6 - 5 \cdot \lambda = 18 - 5\lambda \)

2. \( \begin{vmatrix} 2\lambda & 5 \\ 4 & 6 \end{vmatrix} = 2\lambda \cdot 6 - 5 \cdot 4 = 12\lambda - 20 \)

3. \( \begin{vmatrix} 2\lambda & 3 \\ 4 & \lambda \end{vmatrix} = 2\lambda \cdot \lambda - 3 \cdot 4 = 2\lambda^2 - 12 \)

Substituting these back into the determinant expression:

\[ \Delta = \lambda(18 - 5\lambda) - 2(12\lambda - 20) + 2(2\lambda^2 - 12) \]

Expanding this gives:

\[ \Delta = 18\lambda - 5\lambda^2 - 24\lambda + 40 + 4\lambda^2 - 24 \]

Combining like terms:

\[ \Delta = (-5\lambda^2 + 4\lambda^2) + (18\lambda - 24\lambda) + (40 - 24) \]
\[ \Delta = -\lambda^2 - 6\lambda + 16 \]

### Step 3: Find the values of \( \lambda \) for which \( \Delta = 0 \)
Setting the determinant to zero:

\[ -\lambda^2 - 6\lambda + 16 = 0 \]

Multiplying through by -1:

\[ \lambda^2 + 6\lambda - 16 = 0 \]

Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ \lambda = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \]
\[ = \frac{-6 \pm \sqrt{36 + 64}}{2} \]
\[ = \frac{-6 \pm \sqrt{100}}{2} \]
\[ = \frac{-6 \pm 10}{2} \]

Calculating the two possible values:

1. \( \lambda = \frac{4}{2} = 2 \)
2. \( \lambda = \frac{-16}{2} = -8 \)

### Step 4: Analyze the solutions based on the determinant
- For \( \lambda = 2 \):
- We need to check \( \Delta_1, \Delta_2, \Delta_3 \) to determine if there are no solutions or infinitely many solutions.

Calculating \( \Delta_1 \) (replace the first column with the constants):

\[ \Delta_1 = \begin{vmatrix} 5 & 2 & 2 \\ 8 & 3 & 5 \\ 10 & \lambda & 6 \end{vmatrix} \]

Calculating this determinant gives a non-zero value, indicating no solution.

- For \( \lambda = -8 \):
- \( \Delta \) is zero, indicating a unique solution.

### Conclusion
The correct options based on the analysis are:
- No solution when \( \lambda = 2 \)
- A unique solution when \( \lambda = -8 \)