If the sides AB, BC and CA of a triangle ABC have 3, 5 and 6...

We know that for making a triangle, we require three points, which are not on a line.

Here, we have total $3 + 5 + 6 = 14$ points out of which $3, 5, 6$ points respectively, on $AB, BC$ and $CA$ are collinear.

Thus, the total Number of triangles formed

$= C3 14 - C3 3 - C3 5 - C3 6$

$= \dfrac{14 !}{3 ! \cdot 11 !} - \dfrac{3 !}{3 ! \cdot 0 !} - \dfrac{5 !}{3 ! \cdot 2 !} - \dfrac{6 !}{3 ! \cdot 3 !}$

$= \dfrac{14 \times 13 \times 12 \times 11 !}{3 \times 2 \times 1 \times 11 !} - 1 - \dfrac{5 \times 4 \times 3 !}{3 ! \times 2 \times 1} - \dfrac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !}$

$364 - 1 - 10 - 20 = 333 .$

Hence, total number of triangles is $333 .$