The system of linear equations
`lambdax+2y+2z=5`
`2lambda x+3y+5z=8`
`4x+lambday+6z=10` has :

Options

A. no solution when `lambda`=2
B. infinitely many solutions when `lambda=2`
C. no solution when `lambda=8`
D. infinite solution when `lambda=-8`

Answer

To determine the conditions under which the system of linear equations has no solution or infinitely many solutions, we can analyze the given equations step by step.

### Given Equations:
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)

### Step 1: Write the system in matrix form
We can represent the system of equations in the form of a matrix \( A \) and a vector \( b \):

\[ A = \begin{pmatrix} \lambda & 2 & 2 \\ 2\lambda & 3 & 5 \\ 4 & \lambda & 6 \end{pmatrix}, \quad b = \begin{pmatrix} 5 \\ 8 \\ 10 \end{pmatrix} \]

### Step 2: Find the rank of matrix \( A \)
To find the rank of the matrix \( A \), we will perform row operations to reduce it to row echelon form.

1. Start with the original matrix:
\[ \begin{pmatrix} \lambda & 2 & 2 \\ 2\lambda & 3 & 5 \\ 4 & \lambda & 6 \end{pmatrix} \]

2. Replace Row 2 with Row 2 - 2 * Row 1:
\[ R_2 \rightarrow R_2 - 2R_1 \]
This gives:
\[ \begin{pmatrix} \lambda & 2 & 2 \\ 0 & 3 - 4 & 5 - 4 \\ 4 & \lambda & 6 \end{pmatrix} = \begin{pmatrix} \lambda & 2 & 2 \\ 0 & -1 & 1 \\ 4 & \lambda & 6 \end{pmatrix} \]

3. Replace Row 3 with Row 3 - \( \frac{4}{\lambda} \) * Row 1 (if \( \lambda \neq 0 \)):
\[ R_3 \rightarrow R_3 - \frac{4}{\lambda}R_1 \]
This gives:
\[ \begin{pmatrix} \lambda & 2 & 2 \\ 0 & -1 & 1 \\ 0 & \lambda - \frac{8}{\lambda} & 6 - \frac{8}{\lambda} \end{pmatrix} \]

### Step 3: Analyze the rank conditions
To determine if the system has no solution or infinitely many solutions, we need to check the rank of the augmented matrix \( [A|b] \):

\[ [A|b] = \begin{pmatrix} \lambda & 2 & 2 & | & 5 \\ 2\lambda & 3 & 5 & | & 8 \\ 4 & \lambda & 6 & | & 10 \end{pmatrix} \]

### Step 4: Conditions for no solution or infinitely many solutions
- **No Solution**: This occurs when \( \text{rank}(A) < \text{rank}(A|b) \).
- **Infinitely Many Solutions**: This occurs when \( \text{rank}(A) = \text{rank}(A|b) < \text{number of variables} \).

### Step 5: Calculate specific values of \( \lambda \)
1. If \( \lambda = 2 \):
- Substitute \( \lambda = 2 \) into the matrix and check the ranks.
- The rank of \( A \) becomes 2, while the rank of \( [A|b] \) becomes 3, indicating no solution.

2. If \( \lambda \neq 2 \):
- The ranks may be equal or differ depending on the values of \( \lambda \).

### Conclusion
From the analysis, we find that:
- The system has **no solution** when \( \lambda = 2 \).
- For other values of \( \lambda \), we need to check specific cases to determine if we have infinitely many solutions.

### Final Answer
The system of linear equations has:
- **No solution** for \( \lambda = 2 \).