If $A=(0,0,0)$ and $B=(1,0,0)$ are two points of a line in three dimensions, I think its equation should be $$\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}\tag1$$ according to the formula $$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\tag2$$
But (1) has zeros in the denominator. How to interpret the relation between $x$,$y$ and $z$ then?
On
Your equation doesn't quite work, since it has division by $0$. This is somewhat of an issue. However, think about it logically: $(0,0,0)$ is the origin and $(1,0,0)$ is some point on the $x$-axis. Hey, those are two points on the $x$-axis, which is a line, and hence is the unique line between them. The $x$-axis is clearly defined by $$y=z=0$$ and no constraint is placed on $x$.
We can (informally) read your equation two ways - first, we could "multiply" out by $0$ and cancel with the denominators - which yields $0=y=z$. We could also think that the numerators of $\frac{y-0}0$ and $\frac{z-0}0$ better be $0$, since at least that gives us the indeterminate form $\frac{0}0$ instead of the clearly wrong $\frac{1}0$ or something like that. These aren't super formal ways of thinking about it (but they can be formalized with limits), but they'll yield the right answer when used carefully.
I'll answer in general.. "how to find the equation of a line in $3$D given its two endpoints?". Let $(x_0,y_0,z_0)$ and $(x_1,y_1,z_1)$ two points. The line you're looking for is: $${\bf X}(t) = (x_0,y_0,z_0)+t(x_1-x_0,y_1-y_0,z_1-z_0), \quad t \in \Bbb R.$$ If you just want the segment, consider only $0 \leq t \leq 1$. Notice that ${\bf X}(0) = (x_0,y_0,z_0)$ and ${\bf X}(1) = (x_1,y_1,z_1)$.