Consider the system of equations
`x+y+z=6`
`x+2y+3z=10`
`x+2y+lambdaz =mu`
The system has no solution if

Options

A. `lambda ne 3`
B. `lambda =3, mu =10`
C. `lambda =3, mu ne 10`
D. none of these

Explanation

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

### Given System of Equations:
1. \( x + y + z = 6 \) (Equation 1)
2. \( x + 2y + 3z = 10 \) (Equation 2)
3. \( x + 2y + \lambda z = \mu \) (Equation 3)

### Step 1: Form the Coefficient Matrix and the Augmented Matrix
The coefficient matrix \( A \) and the augmented matrix \( [A|B] \) can be formed as follows:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 10 \\ \mu \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix
To find the condition for no solution, we need to calculate the determinant of the coefficient matrix \( A \) and set it to zero.

\[ \text{det}(A) = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix} \]

### Step 3: Expand the Determinant
Using the first row to expand the determinant:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 2 & 3 \\ 2 & \lambda \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 3 \\ 1 & \lambda \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 3 \\ 2 & \lambda \end{vmatrix} = 2\lambda - 6 \)

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

3. \( \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = 0 \)

Putting it all together:

\[ \text{det}(A) = 1(2\lambda - 6) - 1(\lambda - 3) + 0 \]
\[ = 2\lambda - 6 - \lambda + 3 = \lambda - 3 \]

### Step 4: Set the Determinant to Zero
For the system to have no solution, the determinant must be zero:

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

### Step 5: Calculate the Condition for \( \mu \)
Now we need to ensure that the system does not have a solution when \( \lambda = 3 \). We will calculate the determinant of the augmented matrix \( [A|B] \):

\[ \text{det}(A') = \begin{vmatrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 3 & | & 10 \\ 1 & 2 & 3 & | & \mu \end{vmatrix} \]

Replacing the third column with the constants from the equations:

\[ \text{det}(A') = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 1 & 2 & \mu \end{vmatrix} \]

### Step 6: Expand the Determinant of the Augmented Matrix
Using the first row to expand:

\[ \text{det}(A') = 1 \cdot \begin{vmatrix} 2 & 10 \\ 2 & \mu \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 10 \\ 1 & \mu \end{vmatrix} + 6 \cdot \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} 2 & 10 \\ 2 & \mu \end{vmatrix} = 2\mu - 20 \)

2. \( \begin{vmatrix} 1 & 10 \\ 1 & \mu \end{vmatrix} = \mu - 10 \)

3. \( \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = 0 \)

Putting it all together:

\[ \text{det}(A') = 1(2\mu - 20) - 1(\mu - 10) + 0 \]
\[ = 2\mu - 20 - \mu + 10 = \mu - 10 \]

### Step 7: Set the Condition for \( \mu \)
For the system to have no solution, we need:

\[ \mu - 10 \neq 0 \implies \mu \neq 10 \]

### Conclusion
The system of equations has no solution if:

\[ \lambda = 3 \quad \text{and} \quad \mu \neq 10 \]