There 12 points in a plane of which 5 are collinear . Find
the number of triangles that can be formed with vertices at these points.

Answer

To solve the problem of finding the number of triangles that can be formed with vertices at 12 points in a plane, where 5 of the points are collinear, we can break down the solution into the following steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a total of 12 points, out of which 5 points are collinear. A triangle cannot be formed if all three vertices are collinear. Therefore, we need to consider different cases based on the selection of points.

2. **Case 1: 2 Points from Collinear Points and 1 Point from Non-Collinear Points**:
- We can select 2 points from the 5 collinear points and 1 point from the remaining 7 non-collinear points.
- The number of ways to choose 2 points from 5 is given by the combination formula \( \binom{n}{r} \), which is calculated as \( \frac{n!}{r!(n-r)!} \).
- Thus, the number of ways to choose 2 points from 5 is \( \binom{5}{2} \) and the number of ways to choose 1 point from 7 is \( \binom{7}{1} \).
- Therefore, the total for this case is:
\[ \text{Case 1} = \binom{5}{2} \times \binom{7}{1} \]

3. **Case 2: 1 Point from Collinear Points and 2 Points from Non-Collinear Points**:
- We can select 1 point from the 5 collinear points and 2 points from the 7 non-collinear points.
- The number of ways to choose 1 point from 5 is \( \binom{5}{1} \) and the number of ways to choose 2 points from 7 is \( \binom{7}{2} \).
- Therefore, the total for this case is:
\[ \text{Case 2} = \binom{5}{1} \times \binom{7}{2} \]

4. **Case 3: All Points from Non-Collinear Points**:
- We can select all 3 points from the 7 non-collinear points.
- The number of ways to choose 3 points from 7 is \( \binom{7}{3} \).
- Therefore, the total for this case is:
\[ \text{Case 3} = \binom{7}{3} \]

5. **Total Number of Triangles**:
- The total number of triangles formed is the sum of the triangles from all three cases:
\[ \text{Total} = \text{Case 1} + \text{Case 2} + \text{Case 3} \]
- Substituting the values:
\[ \text{Total} = \left( \binom{5}{2} \times \binom{7}{1} \right) + \left( \binom{5}{1} \times \binom{7}{2} \right) + \left( \binom{7}{3} \right) \]

6. **Calculating the Combinations**:
- Calculate each combination:
- \( \binom{5}{2} = \frac{5!}{2!(5-2)!} = 10 \)
- \( \binom{7}{1} = 7 \)
- \( \binom{5}{1} = 5 \)
- \( \binom{7}{2} = \frac{7!}{2!(7-2)!} = 21 \)
- \( \binom{7}{3} = \frac{7!}{3!(7-3)!} = 35 \)

7. **Final Calculation**:
- Substitute the values back into the total:
\[ \text{Total} = (10 \times 7) + (5 \times 21) + 35 \]
\[ = 70 + 105 + 35 = 210 \]

### Final Answer:
The total number of triangles that can be formed is **210**.