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 product of all 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 third line is either parallel to one of the first two lines or passes through their intersection point.

### Step-by-Step Solution:

1. **Identify the slopes of the given lines:**
- For the line \(x + 2y - 3 = 0\):
\[ 2y = -x + 3 \implies y = -\frac{1}{2}x + \frac{3}{2} \]
The slope is \(-\frac{1}{2}\).

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

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

2. **Set conditions for parallel lines:**
- The third line is parallel to the first line if:
\[ -\lambda = -\frac{1}{2} \implies \lambda = \frac{1}{2} \]
- The third line is parallel to the second line if:
\[ -\lambda = 3 \implies \lambda = -3 \]

3. **Find the intersection point of the first two lines:**
- Solve the equations:
\[ x + 2y - 3 = 0 \quad (1) \]
\[ 3x - y - 1 = 0 \quad (2) \]
- From (1), express \(x\) in terms of \(y\):
\[ x = 3 - 2y \]
- Substitute into (2):
\[ 3(3 - 2y) - y - 1 = 0 \implies 9 - 6y - y - 1 = 0 \implies 8 - 7y = 0 \implies y = \frac{8}{7} \]
- Substitute \(y\) back to find \(x\):
\[ x = 3 - 2\left(\frac{8}{7}\right) = 3 - \frac{16}{7} = \frac{21 - 16}{7} = \frac{5}{7} \]
- The intersection point is \(\left(\frac{5}{7}, \frac{8}{7}\right)\).

4. **Condition for the third line to pass through the intersection point:**
- Substitute the intersection point into the third line:
\[ \lambda\left(\frac{5}{7}\right) + \left(\frac{8}{7}\right) - 2 = 0 \]
Simplifying gives:
\[ \frac{5\lambda}{7} + \frac{8}{7} - 2 = 0 \implies \frac{5\lambda + 8 - 14}{7} = 0 \implies 5\lambda - 6 = 0 \implies \lambda = \frac{6}{5} \]

5. **Values of \(\lambda\) that cannot form a triangle:**
- From the previous steps, we have:
- \(\lambda = \frac{1}{2}\)
- \(\lambda = -3\)
- \(\lambda = \frac{6}{5}\)

6. **Calculate the product of all values of \(|\lambda|\):**
- The absolute values are:
- \(|\frac{1}{2}| = \frac{1}{2}\)
- \(|-3| = 3\)
- \(|\frac{6}{5}| = \frac{6}{5}\)
- The product is:
\[ \frac{1}{2} \times 3 \times \frac{6}{5} = \frac{18}{10} = \frac{9}{5} = 1.8 \]

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