Answer: 272

Small Hint:

If $D$ and $d$ are the common differences of $(A,B,C)$ and $(b,a,c),$ then mod $p$ both $D \equiv -d$ and $D \equiv 2d,$ so $3d$ is divisible by $p$

Big Hint:

Since $0 \lt 2d \lt p,$ that forces $p = 3$ and $(b,a,c) = (0,1,2);$ count triples with $A \equiv 1,$ $B \equiv 0,$ $C \equiv 2 \pmod 3$

Solution:

Let $d$ be the common difference of $(b, a, c),$ so $a - b = c - a = d \gt 0$ and $c = b + 2d \lt p,$ whence $0 \lt 2d \lt p.$ Let $D \gt 0$ be the common difference of $(A, B, C).$ Reducing mod $p,$ we get $D = B - A \equiv b - a = -d$ and $D = C - B \equiv c - b = 2d,$ so $3d$ is divisible by $p.$ Since $0 \lt d \lt p,$ the prime $p$ cannot divide $d,$ so $p = 3;$ then $2d \lt 3$ gives $d = 1$ and $(b, a, c) = (0, 1, 2).$

So the valid triples are exactly the increasing arithmetic progressions in $[0, 99]$ with $A \equiv 1,$ $B \equiv 0,$ $C \equiv 2 \pmod 3.$ Write $A = 1 + 3j$ with $j \ge 0;$ the difference satisfies $D \equiv -1 \equiv 2 \pmod 3,$ so $D = 2 + 3k$ with $k \ge 0.$ The constraint is $C = A + 2D$ $= 5 + 3j + 6k \le 99,$ i.e. $j + 2k \le 31,$ and every such pair $(j, k)$ works.

For each $k = 0, 1, \ldots, 15$ there are $32 - 2k$ choices of $j,$ so $$\begin{aligned} &N = \sum_{k=0}^{15} (32 - 2k) \\ &= 16 \cdot 32 - 2 \cdot \frac{15 \cdot 16}{2} \\ &= 512 - 240 = 272. \end{aligned}$$