We write $a=3^{r}, b=3^{s}$ and $c=3^{t}$ where each of $r, s, t$ is between 1 and 8, inclusive. Since $a \leq b \leq c$, then $r \leq s \leq t$. Next, we note that $\frac{ab}{c}=\frac{3^{r} 3^{s}}{3^{t}}=3^{r+s-t}$, $\frac{ac}{b}=\frac{3^{r} 3^{t}}{3^{s}}=3^{r+t-s}$, and $\frac{bc}{a}=\frac{3^{s} 3^{t}}{3^{r}}=3^{s+t-r}$. Since $t \geq s$, then $r+t-s=r+(t-s) \geq r>0$ and so $\frac{ac}{b}$ is always an integer. Since $t \geq r$, then $s+t-r=s+(t-r) \geq s>0$ and so $\frac{bc}{a}$ is always an integer. Since $\frac{ab}{c}=3^{r+s-t}$, then $\frac{ab}{c}$ is an integer exactly when $r+s-t \geq 0$ or $t \leq r+s$. This means that we need to count the number of triples $(r, s, t)$ where $r \leq s \leq t$, each of $r, s, t$ is an integer between 1 and 8, inclusive, and $t \leq r+s$. Suppose that $r=1$. Then $1 \leq s \leq t \leq 8$ and $t \leq s+1$. If $s=1, t$ can equal 1 or 2. If $s=2, t$ can equal 2 or 3. This pattern continues so that when $s=7, t$ can equal 7 or 8. When $s=8$, though, $t$ must equal 8 since $t \leq 8$. In this case, there are $2 \times 7+1=15$ pairs of values for $s$ and $t$ that work, and so 15 triples $(r, s, t)$. Suppose that $r=2$. Then $2 \leq s \leq t \leq 8$ and $t \leq s+2$. This means that, when $2 \leq s \leq 6, t$ can equal $s, s+1$ or $s+2$. When $s=7, t$ can equal 7 or 8, and when $s=8, t$ must equal 8. In this case, there are $5 \times 3+2+1=18$ triples. Suppose that $r=3$. Then $3 \leq s \leq t \leq 8$ and $t \leq s+3$. This means that, when $3 \leq s \leq 5, t$ can equal $s, s+1, s+2$, or $s+3$. When $s=6,7,8$, there are $3,2,1$ values of $t$, respectively. In this case, there are $3 \times 4+3+2+1=18$ triples. Suppose that $r=4$. Then $4 \leq s \leq t \leq 8$ and $t \leq s+4$. This means that when $s=4$, there are 5 choices for $t$. As in previous cases, when $s=5,6,7,8$, there are $4,3,2,1$ choices for $t$, respectively. In this case, there are $5+4+3+2+1=15$ triples. Continuing in this way, when $r=5$, there are $4+3+2+1=10$ triples, when $r=6$, there are $3+2+1=6$ triples, when $r=7$, there are $2+1=3$ triples, and when $r=8$, there is 1 triple. The total number of triples $(r, s, t)$ is $15+18+18+15+10+6+3+1=86$. Since the triples $(r, s, t)$ correspond with the triples $(a, b, c)$, then the number of triples $(a, b, c)$ is $N=86$.