Lemma: Let $ABC$ be a triangle with the orthic triangle $DEF$. Let $A',B',C'$ be the feet of perpendiculars from the points $A,B,C$ to $FE,FD,DE$ respectively. Then the lines $AA',BB'$ and $CC'$ are concurrent. Proof: The triangles $ABC$ and $AEF$ are similar so $\frac{DB}{DC}=\frac{A'E}{A'F}$. Similarly, $\frac{EC}{EA}=\frac{B'F}{B'D}$ and $\frac{FA}{FB}=\frac{C'D}{C'E}$. Then $\frac{DB}{DC}.\frac{EC}{EA}.\frac{FA}{FB}=\frac{A'E}{A'F}.\frac{B'F}{B'D}.\frac{C'D}{C'E}$ and we know $\frac{DB}{DC}.\frac{EC}{EA}.\frac{FA}{FB}=1$ so $\frac{A'E}{A'F}.\frac{B'F}{B'D}.\frac{C'D}{C'E}=1$. By using Ceva, we get $DA',EB'$ and $FC'$ are concurrent. We also know that $AD,BE$ and $CF$ are concurrent. By using Cevian Nest, we conclude that $AA',BB'$ and $CC'$ are concurrent. Now let's get back to the question. Let $I_A,I_B$ and $I_C$ be the excenters. We know $ABC$ is the orthic triangle of $I_AI_BI_C$. Also, we know $U,V,W$ are the feet of perpendiculars from the points $I_A,I_B,I_C$ to $BC,AC,AB$ respectively. By using the Lemma, we conclude that $I_AU,I_BV$ and $I_CW$ are concurrent $\Rightarrow r_u,r_v$ and $r_w$ are concurrent.

If $I_a$, $I_b$ and $I_c$ are the ex-centers relative to $A$, $B$ and $C$, respectively, and $I$ is the incenter of $\Delta ABC$, we see through angle chasing that $I$ is also the orthocenter of $\Delta I_aI_bI_c$. We can also deduce that $r_U$ is isogonal to $II_a$ with respect to $\Delta I_aI_bI_c$ and analogous for $r_V$ and $r_W$. Let $O$ be the circumcenter of $\Delta I_aI_bI_c$. Since we know that $O$ and $I$ are isogonal conjugates with respect to $\Delta I_aI_bI_c$, we get that $r_U$, $r_V$ and $r_W$ must pass through $O$.

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It is known that $BU = p - c$, $UC = p - b$, and the equivalent for the other sides, in which $a, b, c$ are the lengths of the sides of $ABC$ and $p = \tfrac{a+b+c}{2}$. Let $P = r_u \cap r_v$, $I_C$ be the excenter relative to $C$, and $H$ be the ortocenter of $ABC$. On one hand, the perpendicular projection of the segment $PI_C$ onto the line $CB$ is $b$, and the perpendicular projection of the segment $PI_C$ onto the line $CA$ is $a$. On the other hand, the perpendicular projection of the segment $CH$ onto the line $CB$ is $b\cos C$, and the perpendicular projection of $CH$ onto the line $CA$ is $a\cos C$. Since those form the same ratio, it follows that $AB \perp CH \parallel PI_C$, thus, $PW$ is indeed perpendicular to $AB$ and $P \in r_w$. Therefore, the lines $r_u$, $r_v$ and $r_w$ intersect at $P$.

I wrote this problem and my original solution was just considering the reflection $P$ of $I$ across the circumcenter $O$. Let $R$, $S$, and $T$ be the orthogonal projections of $I$ onto the sides of $ABC$, and $M$, $N$, and $P$ be the orthogonal projections of $O$ onto the sames sides ($M$, $N$, and $P$ are also midpoints of the sides.) Since $U$, $V$, and $W$ are the reflections of $R$, $S$, and $T$ across $M$, $N$, and $P$ respectively (all reflections occurs along a side), $U$, $V$, and $W$ are the orthogonal projections of $P$ onto the sides, and now it is clear that $P$ lies in $r_u = PU$, $r_v = PV$, and $r_w = PW$.

Isn't it well known? Concurrency point is the circumcenter of $I_AI_BI_C$.