To get the answer without proof, this is kind of like an ellipse (in fact, it's a 3-ellipse) with an axis on the $x$-axis. So it's intuitively obvious we'll have extremes when $x=0$ or $y=0$. Below is a proof of the answer.
Particular points
Note that if $x=0$ and $y>0$ we have $12=|z-3|+|z|+|z+3|=|y|+2\sqrt{9+y^2}=y+2\sqrt{9+y^2}$. Thus, $(12-y)^2=4(9+y^2)$ so $144-24y+y^2=36+4y^2$, so that $y^2+8y-36=0$. Combined with $y>0$ we have $y=2(\sqrt{13}-2)<2(\sqrt{16}-2)=4$ so that $|z|<4$ at this point.
And if $y=0$ and $x>0$ we have $12=|x-3|+x+x+3$. If there is a solution with $x\ge3$, then $12=3x$ and we must have $x=4$, which works. So we have $|z|=4$ at this point.
Upper bound and max
Note that the triangle inequality tells us that $12=|z-3|+|z|+|z+3|\ge|z|+|(z-3)+(z+3)|=|z|+|2z|=3|z|$, so that $|z|\le 4$ no matter what. Therefore, $z=4$ attains the maximum $[|z|]=4$.
Lower bound and min
I suspect there is a more elegant way to show the minimum, but I couldn't find one in a reasonable amount of time.
We will show that $12=|z-3|+|z|+|z+3|$ does not intersect the disk $|z|\le3$, by finding where $|z-3|+|z|+|z+3|$ is maximized on the disk and that the maximum value is less than $12$.
Note that the The $y$-partial of $|z-3|+|z|+|z+3|$ is $\dfrac{y}{|z-3|}+\dfrac{y}{|z|}+\dfrac{y}{|z+3|}$ $=y*\dfrac{|z+3||z|+|z-3||z+3|+|z-3||z|}{|z-3||z||z+3|}$. So a potential critical point where this is $0$ or undefined has to have $y=0$ or $z$ at $0$ or $\pm3$ (all of which have $y=0$ anyway), since the three numerator terms are all nonnegative and can't be $0$ simultaneously.
Since we optimize by looking at critical points or the boundary, $|z-3|+|z|+|z+3|$ is maximized over the disk when $y=0$ or when $|z|=3$, In the former case, $|z-3|+|z|+|z+3|=|x|+3-x+x+3=|x|+6\le|z|+6\le3+6=9$. In the latter case, we can use $(x,y)=3(\cos t,\sin t)$ to reduce to a single variable optimization or Lagrange Multipliers. Either way, we find that we need to look at $(\pm3,0)$ and $(0,\pm3)$. We already looked at the points on the $x$-axis, and at $(0,\pm3)$ we have $|z-3|+|z|+|z+3|=2*3\sqrt{2}+3<6*\frac32+3=12$.
Therefore, $|z|\le3$ is impossible, so the lowest possible value of $[|z|]$ is $3$, occurring when $|z|<4$. We found a point with $|z|<4$ already, so the minimum value is indeed $3$.
