Consider the system of equations
`x+y+z=6`
`x+2y+3z=10`
`x+2y+lambdaz =mu`
The system has no solution if (a) `lambda ne 3` (b) `lambda =3, mu =10` (c) `lambda =3, mu ne 10` (d) none of these

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 will analyze the equations 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: Formulate the Coefficient Matrix
The coefficient matrix \( A \) for the system can be represented as:
\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix
To find when the system has no solution, we need to set the determinant of the coefficient matrix to zero:
\[ \text{det}(A) = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix} \]

We can expand this determinant using the first row:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 2 & 3 \\ 2 & \lambda \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 3 \\ 1 & \lambda \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} \]

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

Thus, we have:
\[ \text{det}(A) = (2\lambda - 6) - (\lambda - 3) + 0 \]
\[ = 2\lambda - 6 - \lambda + 3 \]
\[ = \lambda - 3 \]

### Step 3: Set the Determinant to Zero
For the system to have no solution, we set the determinant equal to zero:
\[ \lambda - 3 = 0 \]
\[ \lambda = 3 \]

### Step 4: Analyze the Condition for \( \mu \)
Now we need to check the condition for \( \mu \). If \( \lambda = 3 \), the third equation becomes:
\[ x + 2y + 3z = \mu \]
Now, we compare this with Equation 2:
\[ x + 2y + 3z = 10 \]

For the system to have no solution, the lines represented by these equations must be parallel, which means the ratios of the coefficients must be the same, but the constants must differ:
\[ \frac{10}{\mu} \neq 1 \quad \text{(since the coefficients are the same)} \]
This implies:
\[ 10 \neq \mu \]

### Conclusion
The system of equations has no solution if:
- \( \lambda = 3 \)
- \( \mu \neq 10 \)

Thus, the correct answer is:
**(c) \( \lambda = 3, \mu \neq 10 \)**