The values of `lamda` and` mu` for which the system of equations
`x+y +z=6`
`x+2y +3z = 10`
and `x+2y + lamdaz=mu`
has infinite-number of solutions are .........

Answer

To find the values of \( \lambda \) and \( \mu \) for which the system of equations has an infinite number of solutions, we need to analyze the given equations:

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

For the system to have an infinite number of solutions, the determinant of the coefficients must be zero, and the constants must also satisfy certain conditions.

### Step 1: Form the Coefficient Matrix and Constant Matrix

The coefficient matrix \( A \) and the constant matrix \( B \) can be represented as follows:

\[ A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 10 \\ \mu \end{bmatrix} \]

### Step 2: Calculate the Determinant of the Coefficient Matrix

To find the values of \( \lambda \) such that the determinant of \( A \) is zero, we compute:

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

Using the determinant formula for a 3x3 matrix:

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

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:

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

### Step 3: Set the Determinant to Zero

For the system to have infinite solutions, we set the determinant to zero:

\[ \lambda - 3 = 0 \implies \lambda = 3 \]

### Step 4: Substitute \( \lambda \) Back into the Third Equation

Now, substituting \( \lambda = 3 \) into the third equation:

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

### Step 5: Check for Consistency with the Other Equations

To ensure the system is consistent, we can check if the third equation is a linear combination of the first two equations.

From the first two equations, we can express \( z \) in terms of \( x \) and \( y \):

1. From Equation 1: \( z = 6 - x - y \)
2. Substitute \( z \) into Equation 2:

\[ x + 2y + 3(6 - x - y) = 10 \]

This simplifies to:

\[ x + 2y + 18 - 3x - 3y = 10 \implies -2x - y + 18 = 10 \implies -2x - y = -8 \implies 2x + y = 8 \]

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

Now, substituting \( z \) back into the third equation:

\[ \mu = x + 2y + 3(6 - x - y) = x + 2y + 18 - 3x - 3y = -2x - y + 18 \]

Using \( 2x + y = 8 \):

\[ \mu = 18 - 8 = 10 \]

### Final Answer

Thus, the values of \( \lambda \) and \( \mu \) for which the system of equations has an infinite number of solutions are:

\[ \lambda = 3, \quad \mu = 10 \]