The system (of only one equation)

is compatible and indeterminate, because it has lots of solutions. Indeed, give $x$ and $y$ the values you want, and then solve for $z$. If we choose $x=3$ and $y=1$, then

and $\left(x=3,y=1,z=\frac{15}{4}\right)$ is a solution. If we choose instead $x=0$ and $y=0$, then

and $(x=0,y=0,z=6)$ is a solution. Do you see? Lots of solutions

The system

is compatible and indeterminate, too. It is not so loose though. We may choose the value we like for $x$, but once this choice is fixed, there is no room for other choices. Why? Well, when the value of $x$ is fixed, we are left with something like

for some values $a$ and $b$. But the second row is exactly one $y$ bigger than the first row! So $y$ is necessarily $b-a$. And $z$ would be determined thereafter. Lots of solutions again, but less than in the previous example

The system

is incompatible: it can't be that $2x+3y+4z = 24$ and $2x+3y+4z = 27$ at the same time! The prototype ("minimal example") of incompatible equation, though, is

has exactly one solution, $(x=2,y=8,z=21)$, and therefore is compatible and determinate. But how do we know this? We'll study this in more detail. In the other hand, a system that would be very easily classified as compatible and determinate is