Investigate for what values of `lambda, mu` the simultaneous equations `x + y + z = 6, x + 2 y + 3 z = 10 & x + 2 y + lambda z = mu` have,
A unique solution.

Explanation

To determine the values of \(\lambda\) and \(\mu\) for which the given simultaneous equations have a unique solution, we need to analyze the determinant of the coefficient matrix formed by the equations. The equations are:

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 of equations is:

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

### Step 2: Calculate the Determinant
To find the values of \(\lambda\) and \(\mu\) that allow for a unique solution, we need to compute the determinant of matrix \(A\) and set it not equal to zero.

The determinant of \(A\) is calculated as follows:

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

Using the determinant formula for a \(3 \times 3\) matrix, we can expand it along 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 each of the \(2 \times 2\) 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 these results back into the determinant expression:

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

This simplifies to:

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

### Step 3: Set the Determinant Not Equal to Zero
For the system to have a unique solution, we require:

\[ \lambda - 3 \neq 0 \]

This implies:

\[ \lambda \neq 3 \]

### Step 4: Determine the Values of \(\mu\)
The value of \(\mu\) does not affect the uniqueness of the solution as long as \(\lambda \neq 3\). Therefore, \(\mu\) can take any real number value.

### Final Answer
Thus, the values of \(\lambda\) and \(\mu\) for which the system has a unique solution are:

- \(\lambda \neq 3\)
- \(\mu \in \mathbb{R}\)