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 system of equations given in the problem, we need to analyze the conditions under which the system has more than two solutions. The equations are:

1. \( x + y + z = 6 \) (Equation 1)
2. \( x + 2y + 3z = 10 \) (Equation 2)
3. \( 3x + 2y + \lambda z = \mu \) (Equation 3)

### Step 1: Write the equations in matrix form

We can express the system of equations in matrix form as follows:

\[ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 2 & \lambda \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 10 \\ \mu \end{bmatrix} \]

### Step 2: Determine the condition for infinite solutions

For a system of linear equations to have more than two solutions, the determinant of the coefficient matrix must be zero. This indicates that the equations are dependent.

The determinant of the coefficient matrix is given by:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 2 & \lambda \end{vmatrix} \]

### Step 3: Calculate the determinant

Calculating the determinant \(D\):

\[ D = 1 \begin{vmatrix} 2 & 3 \\ 2 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & 3 \\ 3 & \lambda \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

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

2. \( \begin{vmatrix} 1 & 3 \\ 3 & \lambda \end{vmatrix} = \lambda - 9 \)

3. \( \begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix} = 2 - 6 = -4 \)

Putting it all together:

\[ D = 1(2\lambda - 6) - 1(\lambda - 9) + 1(-4) \]
\[ D = 2\lambda - 6 - \lambda + 9 - 4 \]
\[ D = \lambda + (-1) = \lambda - 1 \]

### Step 4: Set the determinant to zero

For the system to have more than two solutions, we set the determinant to zero:

\[ \lambda - 1 = 0 \]
\[ \lambda = 1 \]

### Step 5: Substitute \(\lambda\) back into the equations

Now, substituting \(\lambda = 1\) back into Equation 3:

\[ 3x + 2y + z = \mu \]

### Step 6: Find \(\mu\)

We need to find \(\mu\) using the solutions we found earlier. We will use the two solutions we derived from the first two equations:

1. First solution: \( (2, 4, 0) \)
2. Second solution: \( (4, 0, 2) \)

Substituting the first solution into Equation 3:

\[ 3(2) + 2(4) + 0 = \mu \implies 6 + 8 = \mu \implies \mu = 14 \]

Substituting the second solution into Equation 3:

\[ 3(4) + 2(0) + 2 = \mu \implies 12 + 0 + 2 = \mu \implies \mu = 14 \]

### Step 7: Calculate \(\mu - \lambda^2\)

Now we need to find \(\mu - \lambda^2\):

\[ \mu - \lambda^2 = 14 - 1^2 = 14 - 1 = 13 \]

### Final Answer

Thus, the value of \(\mu - \lambda^2\) is:

\[ \boxed{13} \]