The product of all the values of `|lambda|`, such that the lines
`x+2y-3=0, 3x-y-1=0` and `lambdax +y-2 =0` cannot form a triangle, is equal to

Answer

To solve the problem, we need to find the values of \(|\lambda|\) such that the lines \(x + 2y - 3 = 0\), \(3x - y - 1 = 0\), and \(\lambda x + y - 2 = 0\) cannot form a triangle. This occurs when either the lines are parallel or all three lines are concurrent.

### Step 1: Determine the slopes of the lines
1. For the line \(L_1: x + 2y - 3 = 0\):
- Rearranging gives \(2y = -x + 3\) or \(y = -\frac{1}{2}x + \frac{3}{2}\).
- The slope \(m_1 = -\frac{1}{2}\).

2. For the line \(L_2: 3x - y - 1 = 0\):
- Rearranging gives \(y = 3x - 1\).
- The slope \(m_2 = 3\).

3. For the line \(L_3: \lambda x + y - 2 = 0\):
- Rearranging gives \(y = -\lambda x + 2\).
- The slope \(m_3 = -\lambda\).

### Step 2: Condition for parallel lines
For lines to be parallel, their slopes must be equal.

1. **Case 1:** Lines \(L_1\) and \(L_3\) are parallel:
\[ m_1 = m_3 \implies -\frac{1}{2} = -\lambda \implies \lambda = \frac{1}{2} \]

2. **Case 2:** Lines \(L_2\) and \(L_3\) are parallel:
\[ m_2 = m_3 \implies 3 = -\lambda \implies \lambda = -3 \]

### Step 3: Condition for concurrent lines
The lines are concurrent if the determinant formed by their coefficients is zero:
\[ \begin{vmatrix} 1 & 2 & -3 \\ 3 & -1 & -1 \\ \lambda & 1 & -2 \end{vmatrix} = 0 \]

Calculating the determinant:
\[ = 1 \begin{vmatrix} -1 & -1 \\ 1 & -2 \end{vmatrix} - 2 \begin{vmatrix} 3 & -1 \\ \lambda & -2 \end{vmatrix} + (-3) \begin{vmatrix} 3 & -1 \\ \lambda & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} -1 & -1 \\ 1 & -2 \end{vmatrix} = (-1)(-2) - (-1)(1) = 2 + 1 = 3\)

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

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

Substituting back into the determinant:
\[ 1(3) - 2(\lambda - 6) - 3(3 + \lambda) = 0 \]
\[ 3 - 2\lambda + 12 - 9 - 3\lambda = 0 \]
\[ 6 - 5\lambda = 0 \implies \lambda = \frac{6}{5} \]

### Step 4: Collecting all values of \(|\lambda|\)
We have found:
- \(\lambda_1 = \frac{1}{2}\)
- \(\lambda_2 = -3\)
- \(\lambda_3 = \frac{6}{5}\)

### Step 5: Calculate the product of \(|\lambda|\)
Calculating the product:
\[ |\lambda_1| \times |\lambda_2| \times |\lambda_3| = \left|\frac{1}{2}\right| \times |-3| \times \left|\frac{6}{5}\right| = \frac{1}{2} \times 3 \times \frac{6}{5} \]
\[ = \frac{18}{10} = \frac{9}{5} \]

### Final Answer
The product of all values of \(|\lambda|\) such that the lines cannot form a triangle is \(\frac{9}{5}\).