Count total available points

The triangle has interior points on its sides:

• On $AB$: $3$ interior points
• On $BC$: $5$ interior points
• On $CA$: $6$ interior points

Including the three vertices $A,B,C$, the total number of points on the boundary is $3+5+6+3=17.$

Count all possible triangles from these 17 points

Any triangle is formed by choosing any $3$ non-collinear points.

First, total ways to choose any $3$ points: $\binom{17}{3} = \frac{17\cdot 16\cdot 15}{6} = 680.$

Subtract collinear triples

The only collinear sets of points lie on the sides $AB$, $BC$, and $CA$.

• On side $AB$, total points on the line are the $3$ interior points plus endpoints $A,B$: $3+2=5.$ Collinear triples on $AB$: $\binom{5}{3}=10.$

• On side $BC$, total points are $5+2=7,$ so collinear triples: $\binom{7}{3}=35.$

• On side $CA$, total points are $6+2=8,$ so collinear triples: $\binom{8}{3}=56.$

Total collinear triples: $10+35+56=101.$

Required number of triangles

$680-101=579.$

Check against options

The computed answer is $579$, which is not present in the given options.

Compare with stored correct answer

Stored correct answer is C: 333, but our correct combinatorial count is $579$.

Hence, the stored answer appears to be incorrect.