$|z-1|+|z-2|+|z-3|=6$ in the Argand Plane In the argand plane  $$C:|z-1|+|z-2|+|z-3|=6 ~~~(1)$$ represents a bounded curve which is a rounded blob  mimicking an "ellipse".
Using the the inequality $$|Z_1+Z_2+Z_3| \le |Z_1|+|Z_2|+|Z_3|, ~~~~(2)$$ we can see that $$6=|z-1|+|z-2|+|z-3|\ge |3z-6| \implies |z-2| \le 2. ~~~(3)$$
The circle $C_1: |z-2|=2$ touches and remains outside  $C$.
Next, it will be interesting to use some other inequality or something to find the equation of a bounded curve $C_2$ which touches and remains inside $C$.  Eventually, $C$ will  touch $C_1$ and $C_2$ and it will be enclosed between them.
The question is to find the curve $C_2.$
Edit:
For $C_2$, the RMS-AM inequality gives:
$$\sqrt{\frac{|z-1|^2+|z-2|^2+|z-3|^2}{3}} \ge \frac{|z-1|+|z-2|+|z-3|}{3}=2~~~~(4)$$
The equality in above doesn't holds as $|z-1|=|z-2|=|z-3|$ cannot be met
$$\implies(x-2)^2+y^2 >\sqrt{\frac{10}{3}}=1.82574~~~~(5)$$
This circle $|z-2|=\sqrt{\frac{10}{3}}$ cannot be the required $C_2$ as  it cannot touch $C$ in (1).

Our claim is that the circle
$$C_2:|z-2|=\sqrt\frac{44}{3}-2$$
lies entirely inside $C$. It is not hard to show that this value is the unique real $r$ for which $2\pm ir$ are on $C$.
We can calculate that
$$|z-2|^2=\frac{|z-1|^2+|z-3|^2}{2}-1.$$
We have that
$$\frac{|z-1|^2+|z-3|^2}{2}\geq\left(\frac{|z-1|+|z-3|}{2}\right)^2,$$
with equality if and only if $|z-1|=|z-3|$, so
$$|z-1|+|z-3|\leq 2\sqrt{|z-2|^2+1},$$
again with equality if and only if $|z-1|=|z-3|$. So, for a point $z$ on $C$,
$$6=|z-1|+|z-2|+|z-3|\leq |z-2|+2\sqrt{|z-2|^2+1};$$
since the right side is increasing in $|z-2|$, this implies that $|z-2|$ is at least the real $r$ for which $6=r+2\sqrt{r^2+1}$, with equality reached if and only if $z=2\pm ir$. It can be calculated that $r=\sqrt\frac{44}3-2$ is the positive value of $r$ that satisfies $6=r+2\sqrt{r^2+1}$, so we are done.
