While out for a stroll, you encounter a vicious velociraptor. You start running away to the northeast at $ 10 \text{m/s}$, and you manage a three-second head start over the raptor. If the raptor runs at $ 15\sqrt{2} \text{m/s}$, but only runs either north or east at any given time, how many seconds do you have until it devours you?
1. **Determine the head start distance:**
   You start running at \(10 \text{ m/s}\) and have a \(3\) second head start. Therefore, the distance you cover in those \(3\) seconds is:
   \[
   \text{Head start distance} = 10 \text{ m/s} \times 3 \text{ s} = 30 \text{ m}
   \]

2. **Calculate the velociraptor's effective speed:**
   The velociraptor runs at \(15\sqrt{2} \text{ m/s}\) but only runs either north or east at any given time. If we assume it splits its time equally between north and east, then in one second, it runs:
   \[
   7.5\sqrt{2} \text{ m/s} \text{ east and } 7.5\sqrt{2} \text{ m/s} \text{ north}
   \]
   Using the Pythagorean theorem, the effective speed in the northeast direction is:
   \[
   \sqrt{(7.5\sqrt{2})^2 + (7.5\sqrt{2})^2} = \sqrt{112.5 + 112.5} = \sqrt{225} = 15 \text{ m/s}
   \]

3. **Calculate the relative speed advantage:**
   You run at \(10 \text{ m/s}\) northeast, and the velociraptor runs at \(15 \text{ m/s}\) northeast. The velociraptor's speed advantage over you is:
   \[
   15 \text{ m/s} - 10 \text{ m/s} = 5 \text{ m/s}
   \]

4. **Determine the time until the velociraptor catches up:**
   The velociraptor needs to make up the \(30 \text{ m}\) head start distance. Given its \(5 \text{ m/s}\) speed advantage, the time it takes to catch up is:
   \[
   \text{Time} = \frac{\text{Head start distance}}{\text{Speed advantage}} = \frac{30 \text{ m}}{5 \text{ m/s}} = 6 \text{ s}
   \]

Conclusion:
\[
\boxed{6 \text{ seconds}}
\]