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

Very sad, isn't it?

The system

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

It seems a stupidity to care about this kind of systems... but it's not! And the reason is that to solve the first system, we will somehow transform it into the second one

which is straightforward to solve. So... ready to know more?