Altered from Another Problem

The expression we want to find is quite well-known and tells exactly what to do. This expression is equal to zero if and only if $N, I, M$ are collinear (Of course you could've just guessed, however, that wouldn't be fun ;) ).

Let $A_1$ be the foot of the angle bisector of angle A on BC.

We will apply $\textbf{Menelaus}$ on $\triangle AA_1D$ .

To finish the problem, it suffices to show that

$\frac{AN}{ND} \cdot \frac{DM}{MA_1} \cdot \frac{A_1I}{IA} = 1$

Obviously, $\frac{AN}{ND} = 1$

We know that $DC = s - c$ , $MC = \frac{a}{2} \implies DM = s - c -\frac{a}{2} = \frac{b-c}{2}$

Also, we know that $\frac{BA_1}{A_1C} = \frac{c}{b} \implies A_1C = \frac{ab}{b+c} \implies MA_1 = \frac{ab}{b+c} - \frac{a}{2} = \frac{a(b-c)}{2(b+c)}$

This implies that $\frac{DM}{MA_1} = \frac{(b+c)}{a}$

By Angle Bisector Theorem on $\triangle ABA_1$ , we find that $\frac{A_1I}{IA} = \frac{a}{b+c}$ (Alternatively you can use mass points, bary, etc. )

Therefore, this implies that

$\frac{AN}{ND} \cdot \frac{DM}{MA_1} \cdot \frac{A_1I}{IA} =1 \cdot \frac{b+c}{a} \cdot \frac{a}{b+c} = 1$

which implies that the three points are collinear.

Therefore, the answer is $\boxed{0}$