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 the lines are either parallel or concurrent.

### Step 1: Convert the lines to slope-intercept form
1. For the first line \( x + 2y - 3 = 0 \):
\[ 2y = -x + 3 \implies y = -\frac{1}{2}x + \frac{3}{2} \]
The slope \( m_1 = -\frac{1}{2} \).

2. For the second line \( 3x - y - 1 = 0 \):
\[ y = 3x - 1 \]
The slope \( m_2 = 3 \).

3. For the third line \( \lambda x + y - 2 = 0 \):
\[ y = -\lambda x + 2 \]
The slope \( m_3 = -\lambda \).

### Step 2: Set conditions for no triangle formation
1. **Condition 1: Lines are parallel**
- For line 3 to be parallel to line 1:
\[ m_3 = m_1 \implies -\lambda = -\frac{1}{2} \implies \lambda = \frac{1}{2} \]

- For line 3 to be parallel to line 2:
\[ m_3 = m_2 \implies -\lambda = 3 \implies \lambda = -3 \implies |\lambda| = 3 \]

2. **Condition 2: Lines are concurrent**
- The lines are concurrent if the point of intersection of lines 1 and 2 lies on line 3.

### Step 3: Find the intersection point of lines 1 and 2
To find the intersection point, we solve the equations:
1. \( x + 2y - 3 = 0 \)
2. \( 3x - y - 1 = 0 \)

From the second equation, we express \( y \):
\[ y = 3x - 1 \]

Substituting into the first equation:
\[ x + 2(3x - 1) - 3 = 0 \implies x + 6x - 2 - 3 = 0 \implies 7x - 5 = 0 \implies x = \frac{5}{7} \]

Now substituting \( x \) back to find \( y \):
\[ y = 3\left(\frac{5}{7}\right) - 1 = \frac{15}{7} - \frac{7}{7} = \frac{8}{7} \]
Thus, the intersection point is \( \left(\frac{5}{7}, \frac{8}{7}\right) \).

### Step 4: Substitute the intersection point into line 3
To find the value of \( \lambda \) such that line 3 passes through the intersection point:
\[ \frac{8}{7} = -\lambda \left(\frac{5}{7}\right) + 2 \]
Multiplying through by 7:
\[ 8 = -5\lambda + 14 \implies -5\lambda = 8 - 14 \implies -5\lambda = -6 \implies \lambda = \frac{6}{5} \]
Thus, \( |\lambda| = \frac{6}{5} \).

### Step 5: Collect all values of \( |\lambda| \)
The values we found are:
1. \( |\lambda| = \frac{1}{2} \)
2. \( |\lambda| = 3 \)
3. \( |\lambda| = \frac{6}{5} \)

### Step 6: Calculate the product of all values of \( |\lambda| \)
\[ \text{Product} = \left(\frac{1}{2}\right) \times 3 \times \left(\frac{6}{5}\right) = \frac{1 \times 3 \times 6}{2 \times 5} = \frac{18}{10} = 1.8 \]

### Final Answer
The product of all the values of \( |\lambda| \) such that the lines cannot form a triangle is \( \boxed{1.8} \).