Let the incircle touch the sides $AB, BC$, and $CA$ at $C_1, A_1$, and $B_1$, respectively. Set $AB = c, BC = a, CA = b$. By equal tangents, we may assume that $AB_1 = AC_1 = x$, $BC_1 = BA_1 = y$, and $CA_1 = CB_1 = z$. Then $a = y + z, b = z + x, c = x + y$. By the AM-GM inequality, we have $a \ge 2\sqrt{yz}$, $b \ge 2\sqrt{zx}$, and $c \ge 2\sqrt{xy}$. Multiplying the last three inequalities yields
$$abc \ge 8xyz, \tag{†}$$
with equality if and only if $x = y = z$; that is, triangle $ABC$ is equilateral.
Let $k$ denote the area of triangle $ABC$. By the Extended Law of Sines, $c = 2R \sin \angle C$. Hence
$$k = \frac{ab \sin \angle C}{2} = \frac{abc}{4R} \quad \text{or} \quad R = \frac{abc}{4k}. \tag{‡}$$
We are going to show that
$$P_A Q_A = \frac{xa^2}{4k}. \tag{*}$$
In exactly the same way, we can also establish its cyclic analogous forms
$$P_B Q_B = \frac{yb^2}{4k} \quad \text{and} \quad P_C Q_C = \frac{zc^2}{4k}.$$
Multiplying the last three equations together gives
$$P_A Q_A \cdot P_B Q_B \cdot P_C Q_C = \frac{xyz a^2 b^2 c^2}{64k^3}.$$
Further considering (†) and (‡), we have
$$8P_A Q_A \cdot P_B Q_B \cdot P_C Q_C = \frac{8xyz a^2 b^2 c^2}{64k^3} \le \frac{a^3 b^3 c^3}{64k^3} = R^3,$$
with equality if and only if triangle $ABC$ is equilateral.
Hence it suffices to show (*). Let $r, r_A, r'_A$ denote the radii of $\omega, \omega_A, \Omega_A$, respectively. We consider the inversion $I$ with center $A$ and radius $x$. Clearly, $I(B_1) = B_1$, $I(C_1) = C_1$, and $I(\omega) = \omega$. Let ray $AO$ intersect $\omega_A$ and $\Omega_A$ at $S$ and $T$, respectively. It is not difficult to see that $AT > AS$, because $\omega$ is tangent to $\omega_A$ and $\Omega_A$ externally and internally, respectively. Set $S_1 = I(S)$ and $T_1 = I(T)$. Let $\ell$ denote the line tangent to $\Omega$ at $A$. Then the image of $\omega_A$ (under the inversion) is the line (denoted by $\ell_1$) passing through $S_1$ and parallel to $\ell$, and the image of $\Omega_A$ is the line (denoted by $\ell_2$) passing through $T_1$ and parallel to $\ell$. Furthermore, since $\omega$ is tangent to both $\omega_A$ and $\Omega_A$, $\ell_1$ and $\ell_2$ are also tangent to the image of $\omega$, which is $\omega$ itself. Thus the distance between these two lines is $2r$; that is, $S_1T_1 = 2r$. Hence we can consider the following configuration. (The darkened circle is $\omega_A$, and its image is the darkened line $\ell_1$.)

By the definition of inversion, we have $AS_1 \cdot AS = AT_1 \cdot AT = x^2$. Note that $AS = 2r_A$, $AT = 2r'_A$, and $S_1T_1 = 2r$. We have
$$r_A = \frac{x^2}{2AS_1}. \quad \text{and} \quad r'_A = \frac{x^2}{2AT_1} = \frac{x^2}{2(AS_1 - 2r)}.$$
Hence
$$P_A Q_A = AQ_A - AP_A = r'_A - r_A = \frac{x^2}{2} \left( \frac{1}{AS_1 - 2r} + \frac{1}{AS_1} \right).$$
Let $H_A$ be the foot of the perpendicular from $A$ to side $BC$. It is well known that $\angle BAS_1 = \angle BAO = 90^\circ - \angle C = \angle CAH_A$. Since ray $AI$ bisects $\angle BAC$, it follows that rays $AS_1$ and $AH_A$ are symmetric with respect to ray $AI$. Further note that both line $l_1$ (passing through $S_1$) and line $BC$ (passing through $H_A$) are tangent to $\omega$. We conclude that $AS_1 = AH_A$. In light of this observation and using the fact $2k = AH_A \cdot BC = (AB + BC + CA)r$, we can compute $P_A Q_A$ as follows:
$$\begin{aligned} P_A Q_A &= \frac{x^2}{2} \left( \frac{1}{AH_A - 2r} - \frac{1}{AH_A} \right) = \frac{x^2}{4k} \left( \frac{2k}{AH_A - 2r} - \frac{2k}{AH_A} \right) \\ &= \frac{x^2}{4k} \left( \frac{1}{\frac{1}{BC} - \frac{2}{AB+BC+CA}} - BC \right) = \frac{x^2}{4k} \left( \frac{1}{\frac{1}{y+z} - \frac{1}{x+y+z}} - (y+z) \right) \\ &= \frac{x^2}{4k} \left( \frac{(y+z)(x+y+z)}{x} - (y+z) \right) \\ &= \frac{x(y+z)^2}{4k} = \frac{xa^2}{4k}, \end{aligned}$$
establishing (*). Our proof is complete.