If the sides A B ,   B C and C A of a triangle A B C have 3 ,   5 and 6 interior points…

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

$={}^{14}\mathrm{C}_3-{}^3\mathrm{C}_3-{}^5\mathrm{C}_3-{}^6\mathrm{C}_3$

$=\frac{14!}{3!·11!}-\frac{3!}{3!·0!}-\frac{5!}{3!·2!}-\frac{6!}{3!·3!}$

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

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

Hence, total number of triangles is $333.$