Given a circle $O$ with radius $R$, and an inscribed acute scalene triangle $ABC$ where $AB$ is the largest side, let $AH_A, BH_B, CH_C$ be the altitudes from $A, B, C$ to $BC, CA, AB$ respectively. Let $D$ be the symmetric point of $H_A$ with respect to $H_BH_C$, and $E$ be the symmetric point of $H_B$ with respect to $H_AH_C$. Let $P$ be the intersection of $AD$ and $BE$, and $H$ be the orthocenter of $\triangle ABC$. We aim to prove that $OP \cdot OH$ is fixed and find this value in terms of $R$.

To solve this, we use complex numbers and the properties of the orthocenter and the circumcircle. Let the circumcircle of $\triangle ABC$ be the unit circle in the complex plane. The orthocenter $H$ of $\triangle ABC$ can be represented as $h = a + b + c$, where $a, b, c$ are the complex numbers corresponding to the vertices $A, B, C$ respectively.

The feet of the altitudes $H_A, H_B, H_C$ can be expressed as:
$$h_a = \frac{1}{2} \left( a + b + c - \frac{bc}{a} \right),$$
and similarly for $h_b$ and $h_c$.

The point $P$, which is the pole of $H$ with respect to the circumcircle, is given by:
$$p = \frac{1}{\overline{h}} = \frac{abc}{ab + bc + ac}.$$

Next, we compute the symmetric points $D$ and $E$. Let $X$ be the foot of the perpendicular from $H_A$ to $H_BH_C$. Solving for $X$ using the properties of perpendiculars in the complex plane, we find:
$$2x = h_a + h_b + a^2 (\overline{h_b - h_a}),$$
which simplifies to:
$$d = 2x - h_a = \frac{1}{2} \left( a + b + c - \frac{ac}{b} - \frac{ab}{c} + \frac{a^3}{bc} \right).$$

We then show that $D, A,$ and $P$ are collinear by computing:
$$\frac{d - a}{\overline{d - a}} = \frac{a - p}{\overline{a - p}} = \frac{a^3 (a + b + c)}{ab + ac + bc}.$$

Finally, since $P$ is the pole of $H$ with respect to the circumcircle, the product $OP \cdot OH$ is given by:
$$OP \cdot OH = R^2.$$

Thus, the value of $OP \cdot OH$ is fixed and equals $R^2$.

The answer is: \boxed{R^2}.