Whether a structure is determinate or indeterminate is just a description of how easy or hard it is to solve.
For a 2D planar structure such as a beam or frame, there are only three global equations at your disposal:
$$\begin{align}
\sum F_x &= 0 \\
\sum F_y &= 0 \\
\sum M_A &= 0
\end{align}$$
These are that the sum of horizontal and vertical forces are null, as is the sum of moments around an arbitrary point $A$. Now, if you have only three reactions, then you only have three unknowns. Since the number of unknowns is equal to the number of equations, then you can "trivially" solve for these reactions with just the equations of static equilibrium above. Therefore, statically-determinate.
Now, if you have more than three reactions, then you have more variables than equations. You can't solve the solution using just the static equilibrium equations. Therefore, statically indeterminate or hyperstatic. For these structures, you need to add more equations which define the compatibility of deformations along your structure.
Consider a continuous beam under a uniform load $q$ with three supports: one at each end and one somewhere along the span (not at mid-span, which would allow cheating), as below:
How much vertical force goes to each support? You have
$$\begin{align}
\sum F_x &= A_x + B_x + C_x = 0 \\
\sum F_y &= A_y + B_y + C_y - q(L_1+L_2) = 0 \\
\sum M_A &= B_yL_1 + C_y(L_1+L_2) - \frac{1}{2}q(L_1+L_2)^2 = 0
\end{align}$$
We can immediately afirm that $A_x = B_x = C_x = 0$. But now we have two equations for three unknowns ($A_y$, $B_y$ and $C_y$). We can use the $\sum F_y$ equation to find $A_y = q(L_1+L_2) - B_y - C_y$. But then we throw that into the $\sum M_A$ equation and get two variables we can't solve for. We're stuck.
We would need to observe that the rotation of segment $\overline{AB}$ at point $B$ is equal to the rotation of segment $\overline{BC}$ at point $B$. Armed with this information, we can create a new equation which describes the compatibility of deformations at point $B$. With this new equation, we now have as many equations as we have unknowns and can therefore solve the problem.
Now, if we didn't have one of those supports (either $A$ or $B$ or $C$), then this problem would be trivial to solve. That is all that is meant by statically determinate or indeterminate. Is it determinate? Then it's simple to solve. Is it indeterminate? Then it's a headache.
Now, if the structure were three-dimensional, then there are six equations of static equilibrium:
$$\begin{align}
\sum F_x &= 0 \\
\sum F_y &= 0 \\
\sum F_z &= 0 \\
\sum M_{A,x} &= 0 \\
\sum M_{A,y} &= 0 \\
\sum M_{A,z} &= 0 \\
\end{align}$$
So, sum of forces in $x$, $y$ and $z$ are null and moments around the $x$, $y$ and $z$ axes for an arbitrary point $A$ are null. So in this case, a statically determinate structure has equation $r = 6N$.
