The system of equations
`lambda x + y + 3z = 0, 2x + mu y - z = 0, 5x + 7y + z = 0 `
has infinitely many solutions in R. Then,

Options

A. `lambda = 2, mu = 3`
B. `lambda = 1, mu = 2 `
C. `lambda = 1, mu = 3 `
D. `lambda = 3, mu = 1`

Answer

To determine the values of \( \lambda \) and \( \mu \) for which the system of equations has infinitely many solutions, we start with the given equations:

1. \( \lambda x + y + 3z = 0 \)
2. \( 2x + \mu y - z = 0 \)
3. \( 5x + 7y + z = 0 \)

### Step 1: Form the Coefficient Matrix
The coefficient matrix \( A \) of the system can be written as:

\[ A = \begin{bmatrix} \lambda & 1 & 3 \\ 2 & \mu & -1 \\ 5 & 7 & 1 \end{bmatrix} \]

### Step 2: Set Up the Determinant
For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero:

\[ \text{det}(A) = 0 \]

### Step 3: Calculate the Determinant
We calculate the determinant using the formula for a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = \lambda \begin{vmatrix} \mu & -1 \\ 7 & 1 \end{vmatrix} - 1 \begin{vmatrix} 2 & -1 \\ 5 & 1 \end{vmatrix} + 3 \begin{vmatrix} 2 & \mu \\ 5 & 7 \end{vmatrix} \]

Calculating each of the \( 2 \times 2 \) determinants:

1. \( \begin{vmatrix} \mu & -1 \\ 7 & 1 \end{vmatrix} = \mu \cdot 1 - (-1) \cdot 7 = \mu + 7 \)

2. \( \begin{vmatrix} 2 & -1 \\ 5 & 1 \end{vmatrix} = 2 \cdot 1 - (-1) \cdot 5 = 2 + 5 = 7 \)

3. \( \begin{vmatrix} 2 & \mu \\ 5 & 7 \end{vmatrix} = 2 \cdot 7 - \mu \cdot 5 = 14 - 5\mu \)

Substituting these back into the determinant calculation:

\[ \text{det}(A) = \lambda(\mu + 7) - 1(7) + 3(14 - 5\mu) \]

### Step 4: Simplify the Determinant
Now we simplify:

\[ \text{det}(A) = \lambda(\mu + 7) - 7 + 42 - 15\mu \]
\[ = \lambda\mu + 7\lambda + 35 - 15\mu \]

### Step 5: Set the Determinant to Zero
Setting the determinant equal to zero gives us the equation:

\[ \lambda\mu + 7\lambda - 15\mu + 35 = 0 \]

### Step 6: Rearranging the Equation
Rearranging the equation:

\[ \lambda\mu + 7\lambda - 15\mu + 35 = 0 \]

### Step 7: Solve for Specific Values
Now, we can test the provided options for \( \lambda \) and \( \mu \):

1. **Option 1**: \( \lambda = 2, \mu = 3 \)
- Substitute: \( 2 \cdot 3 + 7 \cdot 2 - 15 \cdot 3 + 35 \)
- Result: \( 6 + 14 - 45 + 35 = 10 \) (not zero)

2. **Option 2**: \( \lambda = 1, \mu = 2 \)
- Substitute: \( 1 \cdot 2 + 7 \cdot 1 - 15 \cdot 2 + 35 \)
- Result: \( 2 + 7 - 30 + 35 = 14 \) (not zero)

3. **Option 3**: \( \lambda = 1, \mu = 3 \)
- Substitute: \( 1 \cdot 3 + 7 \cdot 1 - 15 \cdot 3 + 35 \)
- Result: \( 3 + 7 - 45 + 35 = 0 \) (this works)

### Final Answer
Thus, the values of \( \lambda \) and \( \mu \) for which the system has infinitely many solutions are:

\[ \lambda = 1, \mu = 3 \]