Solution:

Let $a = BC_0,$ $b = AC_0,$ and $c = AB.$ The altitude gives $C_0C_1 = \frac{ab}{c},$ and $\triangle C_0C_1B \sim \triangle AC_0B$ with ratio $\frac{a}{c}.$ Each later altitude repeats this construction in a triangle scaled by $\frac{a}{c},$ so the segments $C_{n-1}C_n$ form a geometric series and $$\sum_{n=1}^{\infty} C_{n-1}C_n = \frac{ab/c}{1 - a/c} = \frac{ab}{c - a} = 6p = 6(a + b + c).$$

Since $b^2 = c^2 - a^2 = (c - a)(c + a),$ we get $(c - a)(a + b + c) = (c - a)(c + a) + (c - a)b = b(b + c - a),$ so $ab = 6b(b + c - a),$ that is $7a = 6b + 6c.$ Squaring $6c = 7a - 6b$ and using $36c^2 = 36a^2 + 36b^2$ gives $36a^2 = 49a^2 - 84ab,$ hence $13a = 84b.$

Because the side lengths are pairwise relatively prime, $a = 84$ and $b = 13,$ so $c = \frac{7 \cdot 84 - 6 \cdot 13}{6} = 85,$ and indeed $13^2 + 84^2 = 85^2.$ Then $p = 84 + 13 + 85 = 182$ (and $\frac{ab}{c - a} = \frac{84 \cdot 13}{1} = 1092 = 6p$ checks).