On the sides AB, BC, CA of a $$\triangle ABC$$, 3, 4, 5 distinct points (excluding vertices A, B, C) are respectively chosen. The number of triangles that can be constructed using these chosen points as vertices are :

Consider triangle ABC. On side AB, we have chosen 3 distinct points (excluding vertices A and B). On side BC, we have chosen 4 distinct points (excluding vertices B and C). On side CA, we have chosen 5 distinct points (excluding vertices C and A). The total number of points chosen is 3 + 4 + 5 = 12 points.

To form a triangle, we need to select any 3 points from these 12 points. The total number of ways to choose 3 points from 12 is given by the combination formula:

$$\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220$$

However, not every set of 3 points will form a triangle. If the three points are collinear (lie on the same straight line), they cannot form a triangle. Since the points are on the sides of the triangle, collinear points occur only when all three points are chosen from the same side.

We must subtract the cases where three points are collinear. This happens when:

• All three points are on AB: There are 3 points on AB, and we choose all 3. The number of ways is $$\binom{3}{3} = 1$$.
• All three points are on BC: There are 4 points on BC, and we choose any 3. The number of ways is $$\binom{4}{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$$.
• All three points are on CA: There are 5 points on CA, and we choose any 3. The number of ways is $$\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$$.

The total number of collinear triplets is the sum of these cases: 1 + 4 + 10 = 15.

Therefore, the number of triangles is the total number of ways to choose three points minus the collinear triplets:

$$220 - 15 = 205$$

Hence, the correct answer is Option B.