If the system of equations
`x+y+z=2`
`2x+4y-z=6`
`3x+2y+lambdaz=mu`
has infinitely many solutions, then :

Options

A. `2lambda + mu=-14`
B. `lambda + 2mu=14`
C. `lambda - mu=5`
D. `lambda - 2mu= - 5`

Answer

To solve the problem, we need to determine the values of \(\lambda\) and \(\mu\) such that the given system of equations has infinitely many solutions. This occurs when the determinant of the coefficients of the variables is zero, along with the determinants formed by replacing the constant terms.

### Step 1: Write down the equations
The given equations are:
1. \(x + y + z = 2\) (Equation 1)
2. \(2x + 4y - z = 6\) (Equation 2)
3. \(3x + 2y + \lambda z = \mu\) (Equation 3)

### Step 2: Formulate the coefficient matrix and calculate the determinant
The coefficient matrix for the system is:
\[ \begin{bmatrix} 1 & 1 & 1 \\ 2 & 4 & -1 \\ 3 & 2 & \lambda \end{bmatrix} \]

We need to calculate the determinant of this matrix, denoted as \(\Delta\):
\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & -1 \\ 3 & 2 & \lambda \end{vmatrix} \]

### Step 3: Calculate the determinant
Using the determinant formula for a 3x3 matrix:
\[ \Delta = 1 \cdot \begin{vmatrix} 4 & -1 \\ 2 & \lambda \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & -1 \\ 3 & \lambda \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 4 \\ 3 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} 4 & -1 \\ 2 & \lambda \end{vmatrix} = 4\lambda + 2\)

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

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

Now substituting these values back into the determinant:
\[ \Delta = 1(4\lambda + 2) - 1(2\lambda + 3) + 1(-2) \]
\[ = 4\lambda + 2 - 2\lambda - 3 - 2 \]
\[ = 2\lambda - 3 \]

### Step 4: Set the determinant to zero for infinitely many solutions
For the system to have infinitely many solutions:
\[ 2\lambda - 3 = 0 \]
Solving for \(\lambda\):
\[ 2\lambda = 3 \implies \lambda = \frac{3}{2} \]

### Step 5: Calculate the second determinant \(\Delta_1\)
Next, we need to check the determinant formed by replacing the last column with the constants from the equations:
\[ \Delta_1 = \begin{vmatrix} 1 & 1 & 2 \\ 2 & 4 & 6 \\ 3 & 2 & \mu \end{vmatrix} \]

Calculating this determinant:
\[ \Delta_1 = 1 \cdot \begin{vmatrix} 4 & 6 \\ 2 & \mu \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 6 \\ 3 & \mu \end{vmatrix} + 2 \cdot \begin{vmatrix} 2 & 4 \\ 3 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} 4 & 6 \\ 2 & \mu \end{vmatrix} = 4\mu - 12\)

2. \(\begin{vmatrix} 2 & 6 \\ 3 & \mu \end{vmatrix} = 2\mu - 18\)

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

Now substituting these values back into the determinant:
\[ \Delta_1 = 1(4\mu - 12) - 1(2\mu - 18) + 2(-2) \]
\[ = 4\mu - 12 - 2\mu + 18 - 4 \]
\[ = 2\mu + 2 \]

### Step 6: Set the second determinant to zero
For infinitely many solutions:
\[ 2\mu + 2 = 0 \]
Solving for \(\mu\):
\[ 2\mu = -2 \implies \mu = -1 \]

### Step 7: Conclusion
The values of \(\lambda\) and \(\mu\) are:
\[ \lambda = \frac{3}{2}, \quad \mu = -1 \]