Fixing $z_1$ and $z_2$, the locus of $z_3$ for $L \stackrel{def}{=}|z_1-z_2|^2 + |z_2 - z_3|^2 + |z_3 - z_1|^2 = \text{constant}$ are circles centered at mid-point of $z_1,z_2$. For $z_3$ on the circle $|z| = r_3$ to maximize $L$, the locus circle need to tangent to circle $|z| = r_3$ at $z_3$. This implies the origin $0$ lies on the median of $\triangle z_1z_2 z_3$ (a line passing through $z_3$ and mid-point of $z_1,z_2$).
Apply same argument to other two pairs $z_2,z_3$ and $z_3,z_1$,
we find for the configuration where $L$ is maximized, the origin $0$ lies on the intersection of all three medians of $\triangle z_1z_2z_3$. If $\triangle z_1z_2z_3$ is non-degenerate (ie. $z_1$, $z_2$, $z_3$ are not collinear), the intersection is a single point and $0$ is centroid of $\triangle z_1z_2z_3$. In this case, $z_1 + z_2 + z_3 = 0$.
Update
To summarize, in order to maximize $L$, there are only two possible scenarios:
$z_1, z_2, z_3$ are collinear. In this case, it is trivial to see
$L$ is maximized when $z_1,z_2$ is on opposite side of $z_3$ with respect to origin. A configuration that maximize $L$ is $(z_1,z_2,z_3) = (r_1,r_2,-r_3)$ with corresponding
$$\begin{align}
L = L_{1} &\stackrel{def}{=} (r_1+r_3)^2 + (r_2+r_3)^2 + (r_1-r_2)^2\\
&= 3(r_1^2+r_2^2+r_3^2) - (r_1+r_2-r_3)^2\end{align}
$$
Otherwise, $z_1, z_2, z_3$ are not collinear and $z_1 + z_2 + z_3 = 0$ when $L$ is maximized. By a little bit of algebra, we have
$$L = L_{2} \stackrel{def}{=} 3(r_1^2+r_2^2+r_3^2)$$
The million dollar questions is
which scenarios give us the true maximum?
Notice $r_1 < r_2 < r_3$. When $r_3 > r_1 + r_2$. It is impossible to find $z_1, z_2, z_3$ on the circles to satisfy $z_1 + z_2 + z_3 = 0$. This rule out the $2^{nd}$ scenario and maximum $L$ is $L_1$.
On the other direction, let's say $r_3 < r_1 + r_2$. Among all configuration where $z_1, z_2, z_3$ is collinear, $(z_1,z_2,z_3) = (r_1,r_2,-r_3)$
remains to be a configuration that maximizes $L$.
Consider what happens if we keep $z_3$ fixed, increase the imaginary part of $z_1$ and $-z_2$ by a small $\epsilon$. To second order of $\epsilon$, the real part of $z_1$, $z_2$ will be decreased by amounts
$\frac{\epsilon^2}{2r_1}$ and $\frac{\epsilon^2}{2r_2}$ respectively. Plug this into expression of $L$. To second order of $\epsilon$ again, $L$ will be changed by an amount:
$$\begin{align}
\delta L &=
\underbrace{2(r_3+r_1)\frac{-\epsilon^2}{2r_1}}_{\delta((x_1-x_3)^2)}
+
\underbrace{2(r_3+r_2)\frac{-\epsilon^2}{2r_2}}_{\delta((x_2-x_3)^2)}
+
\underbrace{2(r_2-r_1)\left(\frac{\epsilon^2}{r_1} - \frac{\epsilon^2}{r_2}\right)}_{\delta((x_1-x_2)^2)}
+ 6\epsilon^2 + o(\epsilon^2)\\
&= \frac{(r_1+r_2-r_3)(r_2+r_1)}{r_1r_2} \epsilon^2 + o(\epsilon^2)
\end{align}
$$
When $r_1 + r_2 > r_3$, the coefficient of $\epsilon^2$ in $\delta L$ is positive. This means the "maximum" from $1^{st}$ scenario cannot be the true maximum. The true maximum is $L_2$, the one from $2^{nd}$ scenario.
Combine all these, we have
$$\max L = 3(r_1^2+r_2^2+r_3^2) -
\begin{cases}(r_1 + r_2 - r_3)^2, & r_3 > r_1 + r_2\\0, &\text{ otherwise}
\end{cases}$$
No, in fact you can construct such a diagram for any four given positive lengths $R_1,$ $R_2,$ $R_3$ and $R_4.$ First draw $C_1$ with radius $R_1.$ Choose the centre of $C_2$ anywhere on the circle with the same centre as $C_1$ and radius $R_1+R_2.$ Then around those two centres draw circles with radii $R_1+R_4$ resp. $R_2+R_4,$ which must necessarily intersect in two points. Make one of those points the centre of $C_4.$ Finally draw circles around the centres of $C_2$ and $C_4$ with radii $R_2+R_3$ resp. $R_4+R_3.$ Those circles intersect in 2 points on either side of the line connecting their centres: choose the one on the opposite side from the centre of $C_1$ as the centre of $C_3.$
If in addition the angle at $P_2,$ the centre of $C_2$ is given, then $P_4$ and $R_4$ are uniquely determined by the conditions that:
$$d(P_2,P_4)-d(P_1,P_4)=R_2-R_1$$
$$d(P_3,P_4)-d(P_2,P_4)=R_3-R_2$$
$$d(P_3,P_4)-d(P_1,P_4)=R_3-R_1$$
Each of these conditions separately puts $P_4$ on a hyperbola: the locus of points with a given difference in distance to two fixed points. Only 2 out of the 3 conditions are independent, as the third is actually the sum of the first 2. The intersection of these hyperbolae determines $P_4$ and then $R_4$ is the distance from $P_4$ to any of the 3 other circles.
For an actual construction with ruler and compass, have a look at this question and answer.
Best Answer
Let the center of the circle be an origin and $P_i(x_i,y_i).$
Thus, $x_i^2+y_i^2=1$ and $$\sum_{1\leq i<j\leq n}P_iP_j^2=\sum_{1\leq i<j\leq n}\left((x_i-x_j)^2+(y_i-y_j)^2\right)=$$ $$=n(n-1)-2\sum_{1\leq i<j\leq n}(x_ix_j+y_iy_j)=$$ $$= n(n-1)-\left(\sum_{i=1}^nx_i\right)^2-\left(\sum_{i=1}^ny_i\right)^2+\sum_{i=1}^n(x_i^2+y_i^2)\leq$$ $$\leq n(n-1)+n=n^2.$$