$\boxed{\frac{60}{23}}$
Let $\ell$ be the length of a side of square $ABCD$, and $OA = OB = OC = OD = r$, $OP = a$, $OQ = b$, $OR = c$, $OS = d$. Also let $X$, $Y$, $Z$ be the point of intersection of lines $AB$ and $PQ$, lines $BC$ and $QR$, lines $CD$ and $RS$, respectively. Then, by Menelaus' theorem, we have
$$\frac{OP}{AP} \cdot \frac{AX}{BX} \cdot \frac{BQ}{OQ} = \frac{a}{r-a} \cdot \frac{BX + \ell}{BX} \cdot \frac{r-b}{b} = 1,$$
from which we obtain $BX = \frac{\ell a (r-b)}{r(b-a)}$. Similarly, we get $BY = \frac{\ell c (r-b)}{r(b-c)}$, $CZ = \frac{\ell d (r-c)}{r(c-d)}$. Also, we have $CY = BC + BY = \ell + \frac{\ell c (r-b)}{r(b-c)} = \frac{\ell b (r-c)}{r(b-c)}$. Since triangles $YBX$ and $YCZ$ are similar, we get
$$\begin{align*} & BY \cdot CZ = BX \cdot CY \\ \iff & \frac{\ell c (r-b)}{r(b-c)} \cdot \frac{\ell d (r-c)}{r((c-d))} = \frac{\ell a (r-b)}{r(b-a)} \cdot \frac{\ell b (r-c)}{r(b-c)} \\ \iff & cd(b-a) = ab(c-d). \end{align*}$$

Solving for $d$ from the last equation above, we get $d = \frac{abc}{ab+bc-ca}$, and substituting the values $a = 3$, $b = 5$, $c = 4$, we get the length of the line segment $OS$ to be equal to $d = \frac{3 \cdot 4 \cdot 5}{3 \cdot 5 + 5 \cdot 4 - 4 \cdot 3} = \frac{60}{23}$.