Here is a little lemma for $\mathbb{R}^2$ and $k=3$.
Let $c_1,c_2,c_3$ be the centers of disks $D_1,D_2,D_3$, and let $c$ be the center of the enclosing disk $D$, at the center of gravity of the $D_i$'s. I will assume all weights are strictly positive: $w_i > 0$. Define two disks as touching if their circular boundaries touch: from the inside for $D_i$ touching $D$, and externally for $D_i$ touching $D_j$. Define the touching graph $G$ to record which disk touches which.
Claim. In a tightest balanced packing for $k=3$, the touching graph is connected.
First, it is easy to see that the enclosing disk $D$ must touch some $D_i$. For if not, $D$'s radius could be reduced while keeping $c$ fixed. Henceforth, let $D$ touch $D_1$.
Note that $c$ must lie inside $\triangle c_1, c_2, c_3$, and strictly inside (if the triangle has positive area) because $w_i > 0$.
Suppose, in contradiction to the claim, that a disk $D_3$ is not touching any of $\{D, D_1, D_2\}$. Consider now two cases.
(1) The line $L_{12}$ through $c_1$ and $c_2$ contains $c$. Then because $c \in \triangle c_1, c_2, c_3$, it must be that $c_3$ lies on $L_{12}$ as well, and $\triangle$ is a line segment. Then the problem essentially reduces to $\mathbb{R}^1$, where it was earlier established that all disks touch along that line.
(2) $c$ does not lie on $L_{12}$. ($D_2$ may touch $D$ and/or $D_1$.) Then $c_3$ lies on the other side of the diameter of $D$ parallel to $L_{12}$. Move $c_3$ to $c'_3$ toward and perpendicular to $L_{12}$. This moves the center of gravity $c$ to $c'$ in the same direction. See the figure. Recenter $D$ to $D'$ centered on $c'$. Now none of the disks touch $D'$, and so its radius may be reduced.
[![PackingD123][1]][1]