Given the other (incorrect) answers, I think it is worth to give a detailed proof of the following partial negative answer: if the maps $f_i$ are only allowed to be homotheties, then there is no such open set $O$ if $\max(r_1+r_2,r_2+r_3,r_3+r_1)>1$ (which is certainly possible, for example take $r_1=r_2=0.6; r_3=0.1$).

In fact, I claim the following.

*Fact*: already for two numbers $r_1,r_2$ such that $r_1+r_2>1$, it is not possible to find homotheties $f_1, f_2$ of $\mathbb{R}^2$ with ratios $r_1,r_2$, and a nonempty open set $O\subset\mathbb{R}^2$ such that $f_1(O)\subset O$, $f_2(O)\subset O$ and $f_1(O)\cap f_2(O)=\varnothing$. This clearly implies the claim above.

In order to do this I recall some notation and facts. Given a finite collection of strictly contractive similarity maps $g_1,\ldots, g_m$, $m\ge 2$, on $\mathbb{R}^2$, there exists a unique nonempty compact set $E$ such that $E=\bigcup_{i=1}^m g_i(E)$. This set is called the *self-similar set* associated to the *iterated function system* (IFS) $g_1,\ldots,g_m$.

We say that an IFS $(g_1,\ldots,g_m)$ satisfies the *Open Set Condition* (OSC) if there exists a nonempty open set $O$ such that $g_i(O)\subset O$ for $i=1,\ldots,m$ and the $g_i(O)$ are pairwise disjoint.

Note that with this (standard) terminology, the question in the OP can be reformulated as follows: does there exist an IFS $(g_1,g_2,g_3)$ satisfying the OSC where the similarity ratio of $g_i$ is $r_i$?

The *similarity dimension* of the IFS $(g_1,\ldots,g_m)$ is the unique real number $s$ such that $\sum_{i=1}^m r_i^s=1$, where $r_i$ is the similarity ratio of $g_i$.

We then have the following classical result due to Hutchinson (a proof can be found in e.g. Falconer's textbook "Fractal Geometry: mathematical foundations and applications", Chapter 9).

**Theorem**. If $(g_1,\ldots,g_m)$ satisfies the open set condition, then the Hausdorff dimension of the associated self-similar set $E$ equals the similarity dimension $s$.

We can now prove the stated fact. Let $r_1,r_2\in (0,1)$ be such that $r_1+r_2>1$. Let $f_1,f_2$ be homotheties with contraction ratio $r_1,r_2$. Since $r_1+r_2>1$, the similarity dimension of the IFS $(f_1,f_2)$ is strictly larger than $1$.

On the other hand, if we let $p_1,p_2$ be the fixed points of $f_1,f_2$ and $E$ is the closed segment joining them, then one can check directly that $E=f_1(E)\cup f_2(E)$, or in other words the self-similar set associated to $(f_1,f_2)$ is $E$. But $E$ has Hausdorff dimension $1$ (or $0$ when $p_1=p_2$) which is strictly less than the similarity dimension, so the open set condition cannot hold. This is exactly the claimed fact.