Solution:

If $a_{k-1} = a_k$ then $a_{k+2} = a_{k+1}|a_k - a_{k-1}| = 0,$ and if $|a_k - a_{k-1}| = 1$ then $a_{k+2} = a_{k+1},$ so $a_{k+4} = 0.$ Hence every triple of one of the forms $(j,j,k),$ $(j,k,k),$ $(j,j\pm1,k),$ $(j,k,k\pm1)$ produces a $0.$ These forms contain $100 + 100 + 4 \cdot 90 = 560$ triples, but triples fitting two forms are counted twice: the $10$ of the form $(j,j,j),$ the $9$ in each of the six families $(j,j,j\pm1),$ $(j,j\pm1,j),$ $(j,j\pm1,j\pm1)$ (matching signs), and the $8$ in each of $(j,j+1,j+2)$ and $(j,j-1,j-2).$ That leaves $560 - 10 - 54 - 16 = 480$ triples.

A few other triples also work: if $(a_1, a_2, a_3) = (j, j\pm2, 1),$ then $a_4 = 2$ and $|a_4 - a_3| = 1,$ so $a_8 = 0.$ These $16$ triples include $(3,1,1)$ and $(4,2,1),$ which were already counted, so they add $14$ new ones, for $480 + 14 = 494.$

No other triple reaches $0:$ if both consecutive differences are at least $2$ and $a_3 \ge 2,$ then $a_4 = a_3|a_2 - a_1| \ge 2a_3 \gt a_3$ and $|a_4 - a_3| \ge a_3 \ge 2,$ so inductively the terms grow forever and no factor ever vanishes. If instead $a_3 = 1$ with $|a_2 - a_1| \ge 3,$ then $a_4 \ge 3$ and $|a_4 - a_3| \ge 2,$ and the same growth takes over. The count is $494.$