$$\frac{x}2=y=\frac{z}{-2}$$
What is the length of the segment inside of the unit cube?
I guess I should find the intersections of the line and the $x=1,y=1,z=1$ planes but I think this line doesn't pass through the inside of the unit cube. Am I wrong?
It depends on the definition of "unit cube", but let's go with what you proposed in the comments, i.e., the cube bounded by the planes $x=\pm\frac12, y=\pm\frac12, z=\pm\frac12$.
A point lies on the surface of this cube if and only if the absolute value of its largest coördinate is $\frac12$. Therefore the line segment in question is from $(\frac12,\frac14,-\frac12)$ to $(-\frac12,-\frac14,\frac12)$. The point $(1,\frac12,-1)$, in contrast, is on the line, but outside the cube because it is on the wrong side of the planes $x=\frac12$ and $z=-\frac12$.
It can be helpful for problems like this to solve a similar problem in a lower dimension: try finding the length of the line segment in the unit square along the line $\frac x2=y$, for example. You can draw a diagram much more easily this way and see how it relates to the equations.