If system of equation x + y + z = 6 ,x + 2y + 3z = 10, 3x + 2y + `lambda`z = `mu` has more than two solutions. Find `(mu -lambda^2 )`

Explanation

To solve the given system of equations and find the value of \( \mu - \lambda^2 \), we will follow these steps:

### Step 1: Write the system of equations
The given system of equations is:
1. \( x + y + z = 6 \) (Equation 1)
2. \( x + 2y + 3z = 10 \) (Equation 2)
3. \( 3x + 2y + \lambda z = \mu \) (Equation 3)

### Step 2: Form the augmented matrix
We can represent the system of equations in an augmented matrix form:
\[ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 1 & 2 & 3 & | & 10 \\ 3 & 2 & \lambda & | & \mu \end{bmatrix} \]

### Step 3: Perform row operations
We will perform row operations to simplify the augmented matrix.

1. Subtract the first row from the second row:
\[ R_2 = R_2 - R_1 \implies R_2 = [0, 1, 2 | 4] \]

2. Subtract 3 times the first row from the third row:
\[ R_3 = R_3 - 3R_1 \implies R_3 = [0, -1, \lambda - 3 | \mu - 18] \]

So the augmented matrix now looks like:
\[ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 0 & -1 & \lambda - 3 & | & \mu - 18 \end{bmatrix} \]

### Step 4: Further simplify the matrix
Next, we add the second row to the third row:
\[ R_3 = R_3 + R_2 \implies R_3 = [0, 0, \lambda - 1 | \mu - 14] \]

The augmented matrix now is:
\[ \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 2 & | & 4 \\ 0 & 0 & \lambda - 1 & | & \mu - 14 \end{bmatrix} \]

### Step 5: Set conditions for infinite solutions
For the system to have infinitely many solutions, the last row must represent the equation \( 0 = 0 \). Therefore, we set:
1. \( \lambda - 1 = 0 \) (which gives \( \lambda = 1 \))
2. \( \mu - 14 = 0 \) (which gives \( \mu = 14 \))

### Step 6: Calculate \( \mu - \lambda^2 \)
Now we can calculate \( \mu - \lambda^2 \):
\[ \mu - \lambda^2 = 14 - 1^2 = 14 - 1 = 13 \]

### Final Answer
Thus, the value of \( \mu - \lambda^2 \) is \( \boxed{13} \).