1. Define the problem and setup:
Let $ABCD$ be a square with the point of intersection $O$ of the diagonals. Let $P, Q, R, S$ be points on the segments $OA, OB, OC, OD$ respectively such that $OP = 3$, $OQ = 5$, $OR = 4$. We need to find the length of the segment $OS$ given that the points of intersection of the lines $AB$ and $PQ$, $BC$ and $QR$, $CD$ and $RS$ are collinear.

2. Apply Menelaus' Theorem:
We will use Menelaus' theorem on triangle $AOB$ with transversal $PQX$, where $X$ is the intersection of $AB$ and $PQ$.

Menelaus' theorem states that for a triangle $ABC$ with a transversal intersecting $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively, the following holds:
$$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1$$

3. **Apply Menelaus' Theorem to $\triangle AOB$:**
Let $AB = 2l$ and $O$ be the center of the square. The coordinates of $A, B, C, D$ are $(l, l), (l, -l), (-l, -l), (-l, l)$ respectively. The coordinates of $O$ are $(0, 0)$.

For $P$ on $OA$, $Q$ on $OB$, and $X$ on $AB$:
$$OP = 3, \quad OQ = 5, \quad OR = 4$$

Using Menelaus' theorem on $\triangle AOB$ with points $P, Q, X$:
$$\frac{AP}{PO} \cdot \frac{OQ}{QB} \cdot \frac{BX}{XA} = 1$$

Since $AP = l - 3$, $PO = 3$, $OQ = 5$, and $QB = l - 5$:
$$\frac{l - 3}{3} \cdot \frac{5}{l - 5} \cdot \frac{BX}{XA} = 1$$

4. **Solve for $BX/XA$:**
$$\frac{l - 3}{3} \cdot \frac{5}{l - 5} = \frac{BX}{XA}$$
$$\frac{(l - 3) \cdot 5}{3 \cdot (l - 5)} = \frac{BX}{XA}$$
$$\frac{5(l - 3)}{3(l - 5)} = \frac{BX}{XA}$$

5. **Apply Menelaus' Theorem to $\triangle BOC$:**
Similarly, apply Menelaus' theorem to $\triangle BOC$ with points $Q, R, Y$ where $Y$ is the intersection of $BC$ and $QR$:
$$\frac{BQ}{QO} \cdot \frac{OR}{RC} \cdot \frac{CY}{YB} = 1$$

Since $BQ = l - 5$, $QO = 5$, $OR = 4$, and $RC = l - 4$:
$$\frac{l - 5}{5} \cdot \frac{4}{l - 4} \cdot \frac{CY}{YB} = 1$$

6. **Solve for $CY/YB$:**
$$\frac{l - 5}{5} \cdot \frac{4}{l - 4} = \frac{CY}{YB}$$
$$\frac{4(l - 5)}{5(l - 4)} = \frac{CY}{YB}$$

7. **Apply Menelaus' Theorem to $\triangle COD$:**
Finally, apply Menelaus' theorem to $\triangle COD$ with points $R, S, Z$ where $Z$ is the intersection of $CD$ and $RS$:
$$\frac{CR}{RO} \cdot \frac{OS}{SD} \cdot \frac{DZ}{ZC} = 1$$

Since $CR = l - 4$, $RO = 4$, $OS = x$, and $SD = l - x$:
$$\frac{l - 4}{4} \cdot \frac{x}{l - x} \cdot \frac{DZ}{ZC} = 1$$

8. **Solve for $DZ/ZC$:**
$$\frac{l - 4}{4} \cdot \frac{x}{l - x} = \frac{DZ}{ZC}$$

9. Collinearity Condition:
Given that the points of intersection are collinear, we use the condition that the ratios must be equal:
$$\frac{5(l - 3)}{3(l - 5)} = \frac{4(l - 5)}{5(l - 4)} = \frac{(l - 4)x}{4(l - x)}$$

10. **Solve for $x$:**
Equate the first two ratios:
$$\frac{5(l - 3)}{3(l - 5)} = \frac{4(l - 5)}{5(l - 4)}$$
Cross-multiply and solve for $l$:
$$25(l - 3)(l - 4) = 12(l - 5)^2$$
Expand and simplify:
$$25(l^2 - 7l + 12) = 12(l^2 - 10l + 25)$$
$$25l^2 - 175l + 300 = 12l^2 - 120l + 300$$
$$13l^2 - 55l = 0$$
$$l(13l - 55) = 0$$
Since $l \neq 0$, we have:
$$l = \frac{55}{13}$$

Now, substitute $l = \frac{55}{13}$ into the third ratio:
$$\frac{4(l - 5)}{5(l - 4)} = \frac{(l - 4)x}{4(l - x)}$$
$$\frac{4(\frac{55}{13} - 5)}{5(\frac{55}{13} - 4)} = \frac{(\frac{55}{13} - 4)x}{4(\frac{55}{13} - x)}$$
Simplify:
$$\frac{4(\frac{55 - 65}{13})}{5(\frac{55 - 52}{13})} = \frac{(\frac{55 - 52}{13})x}{4(\frac{55}{13} - x)}$$
$$\frac{4(-\frac{10}{13})}{5(\frac{3}{13})} = \frac{(\frac{3}{13})x}{4(\frac{55}{13} - x)}$$
$$\frac{-40}{15} = \frac{3x}{4(\frac{55}{13} - x)}$$
$$-\frac{8}{3} = \frac{3x}{4(\frac{55}{13} - x)}$$
Cross-multiply and solve for $x$:
$$-8 \cdot 4(\frac{55}{13} - x) = 9x$$
$$-32(\frac{55}{13} - x) = 9x$$
$$-32 \cdot \frac{55}{13} + 32x = 9x$$
$$-\frac{1760}{13} + 32x = 9x$$
$$32x - 9x = \frac{1760}{13}$$
$$23x = \frac{1760}{13}$$
$$x = \frac{1760}{13 \cdot 23}$$
$$x = \frac{1760}{299}$$
$$x = \frac{60}{23}$$

The final answer is $\boxed{\frac{60}{23}}$.