Consider the system of equations
`{:(2x+lambday+6z=8),(x+2y+muz=5),(x+y+3z=4):}`
The system of equations has :
No solution if :

Options

A. 1)`lambda=2, mu=3`
B. 2)`lambdane 2 , mu=3`
C. 3)`lambda ne 2, mu ne 3 `
D. 4)`lambda =2 , mu in R `

Answer

To determine the conditions under which the given system of equations has no solution, we will follow these steps:

### Step 1: Write the System of Equations
The given system of equations is:
1. \( 2x + \lambda y + 6z = 8 \)
2. \( x + 2y + \mu z = 5 \)
3. \( x + y + 3z = 4 \)

### Step 2: Form the Coefficient Matrix and Calculate the Determinant
The coefficient matrix \( A \) is:
\[ A = \begin{bmatrix} 2 & \lambda & 6 \\ 1 & 2 & \mu \\ 1 & 1 & 3 \end{bmatrix} \]

To find the condition for no solution, we need to calculate the determinant \( \Delta \) of this matrix and set it to zero:
\[ \Delta = \begin{vmatrix} 2 & \lambda & 6 \\ 1 & 2 & \mu \\ 1 & 1 & 3 \end{vmatrix} \]

### Step 3: Expand the Determinant
We can expand the determinant along the first row:
\[ \Delta = 2 \begin{vmatrix} 2 & \mu \\ 1 & 3 \end{vmatrix} - \lambda \begin{vmatrix} 1 & \mu \\ 1 & 3 \end{vmatrix} + 6 \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 2 & \mu \\ 1 & 3 \end{vmatrix} = (2)(3) - (1)(\mu) = 6 - \mu \)
2. \( \begin{vmatrix} 1 & \mu \\ 1 & 3 \end{vmatrix} = (1)(3) - (1)(\mu) = 3 - \mu \)
3. \( \begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} = (1)(1) - (1)(2) = 1 - 2 = -1 \)

Substituting these back into the determinant:
\[ \Delta = 2(6 - \mu) - \lambda(3 - \mu) + 6(-1) \]
\[ = 12 - 2\mu - 3\lambda + \lambda\mu - 6 \]
\[ = \lambda\mu - 3\lambda - 2\mu + 6 \]

### Step 4: Set the Determinant to Zero
For the system to have no solution, we set \( \Delta = 0 \):
\[ \lambda\mu - 3\lambda - 2\mu + 6 = 0 \]

### Step 5: Rearranging the Equation
Rearranging the equation gives:
\[ \lambda\mu - 3\lambda - 2\mu + 6 = 0 \]
This can be factored or analyzed to find conditions on \( \lambda \) and \( \mu \).

### Step 6: Analyze the Conditions
From the determinant condition, we can see that:
1. If \( \lambda = 2 \), then the equation simplifies to \( 2\mu - 6 = 0 \) which gives \( \mu = 3 \).
2. If \( \mu = 3 \), then the equation simplifies to \( \lambda(3) - 3\lambda - 6 = 0 \) which gives \( \lambda = 2 \).

### Step 7: Check the Second Condition
For the system to have no solution, at least one of the determinants \( \Delta_x, \Delta_y, \Delta_z \) must be non-zero when \( \Delta = 0 \).

### Conclusion
The system of equations has no solution if:
- \( \lambda \neq 2 \) and \( \mu = 3 \).

Thus, the correct answer is:
**The system has no solution if \( \lambda \neq 2 \) and \( \mu = 3 \).**