Find all values of `lambda` for which the `(lambda-1)x+(3lambda+1)y+2lambdaz=0(lambda-1)x+(4lambda-2)y+(lambda+3)z=0 2x+(3lambda+1)y+3(lambda-1)z=0` possess non-trivial solution and find the ratios `x:y:z,` where `lambda` has the smallest of these value.

Options

A. `3 : 2 :1`
B. `3 : 3 : 2`
C. `1 : 3 : 1`
D. `1 : 1 : 1`

Explanation

To solve the problem, we need to find values of \( \lambda \) for which the following system of equations has a non-trivial solution:

1. \((\lambda - 1)x + (3\lambda + 1)y + 2\lambda z = 0\)
2. \((\lambda - 1)x + (4\lambda - 2)y + (\lambda + 3)z = 0\)
3. \(2x + (3\lambda + 1)y + 3(\lambda - 1)z = 0\)

### Step 1: Form the Coefficient Matrix
We can express the system in matrix form as follows:

\[ \begin{bmatrix} \lambda - 1 & 3\lambda + 1 & 2\lambda \\ \lambda - 1 & 4\lambda - 2 & \lambda + 3 \\ 2 & 3\lambda + 1 & 3(\lambda - 1) \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0 \]

### Step 2: Set Up the Determinant
For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero:

\[ D = \begin{vmatrix} \lambda - 1 & 3\lambda + 1 & 2\lambda \\ \lambda - 1 & 4\lambda - 2 & \lambda + 3 \\ 2 & 3\lambda + 1 & 3(\lambda - 1) \end{vmatrix} = 0 \]

### Step 3: Calculate the Determinant
We can calculate the determinant using cofactor expansion.

\[ D = (\lambda - 1) \begin{vmatrix} 4\lambda - 2 & \lambda + 3 \\ 3\lambda + 1 & 3(\lambda - 1) \end{vmatrix} - (3\lambda + 1) \begin{vmatrix} \lambda - 1 & \lambda + 3 \\ 2 & 3(\lambda - 1) \end{vmatrix} + 2\lambda \begin{vmatrix} \lambda - 1 & 4\lambda - 2 \\ 2 & 3\lambda + 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

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

### Step 4: Solve the Determinant Equation
After calculating the determinants and simplifying, we set \( D = 0 \) and solve for \( \lambda \). This will yield values for \( \lambda \).

### Step 5: Find the Ratios \( x:y:z \)
Once we have the values of \( \lambda \), we substitute the smallest value of \( \lambda \) back into the original equations to find the ratios \( x:y:z \).