Consider the system of equations
`x+y+z=6`
`x+2y+3z=10`
`x+2y+lambdaz =mu`
the system has unique 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 a unique solution, we need to analyze the coefficient matrix of the system and ensure that its determinant is non-zero.

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

### Step 1: Form the Coefficient Matrix
The coefficient matrix \( A \) for the system can be expressed as follows:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix
To find the determinant of matrix \( A \), we can use the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix:

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

Calculating each term:

1. First term: \( 1 \cdot (2\lambda - 6) = 2\lambda - 6 \)
2. Second term: \( -1 \cdot (\lambda - 3) = -\lambda + 3 \)
3. Third term: \( 1 \cdot (2 - 2) = 0 \)

Combining these results:

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

Simplifying:

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

### Step 3: Set the Determinant Not Equal to Zero
For the system to have a unique solution, the determinant must be non-zero:

\[ \lambda - 3 \neq 0 \]

This implies:

\[ \lambda \neq 3 \]

### Conclusion
The system of equations has a unique solution if \( \lambda \neq 3 \).

### Answer
The correct option is (a) \( \lambda \neq 3 \).