1. **State the problem:** We have points $P(2,1,3)$, $Q(4,3,4)$, and $R(2,1,8)$. Point $N$ is the foot of the perpendicular from $P$ to the line joining $Q$ and $R$. We need to find the distance from the origin $O(0,0,0)$ to point $N$. 2. **Formula and approach:** The foot of the perpendicular $N$ from point $P$ to line $QR$ can be found using vector projection. Let vector $\vec{QR} = R - Q = (2-4,1-3,8-4) = (-2,-2,4)$. Let vector $\vec{QP} = P - Q = (2-4,1-3,3-4) = (-2,-2,-1)$. The parameter $t$ for the projection of $\vec{QP}$ onto $\vec{QR}$ is given by: $$t = \frac{\vec{QP} \cdot \vec{QR}}{\vec{QR} \cdot \vec{QR}}$$ 3. **Calculate dot products:** $$\vec{QP} \cdot \vec{QR} = (-2)(-2) + (-2)(-2) + (-1)(4) = 4 + 4 - 4 = 4$$ $$\vec{QR} \cdot \vec{QR} = (-2)^2 + (-2)^2 + 4^2 = 4 + 4 + 16 = 24$$ 4. **Calculate $t$:** $$t = \frac{4}{24} = \frac{1}{6}$$ 5. **Find coordinates of $N$:** $$N = Q + t \vec{QR} = (4,3,4) + \frac{1}{6}(-2,-2,4) = \left(4 - \frac{2}{6}, 3 - \frac{2}{6}, 4 + \frac{4}{6}\right) = \left(4 - \frac{1}{3}, 3 - \frac{1}{3}, 4 + \frac{2}{3}\right) = \left(\frac{11}{3}, \frac{8}{3}, \frac{14}{3}\right)$$ 6. **Distance from origin to $N$:** $$ON = \sqrt{\left(\frac{11}{3}\right)^2 + \left(\frac{8}{3}\right)^2 + \left(\frac{14}{3}\right)^2} = \frac{1}{3} \sqrt{11^2 + 8^2 + 14^2} = \frac{1}{3} \sqrt{121 + 64 + 196} = \frac{1}{3} \sqrt{381}$$ 7. **Final answer:** $$\boxed{\frac{\sqrt{381}}{3}}$$