1. **Problem statement:** Determine if the point $P(1|3|0)$ lies on the line passing through points $A(3|2|0)$ and $B(-1|4|0)$. 2. **Formula and concept:** The vector equation of a line through points $A$ and $B$ is given by: $$\vec{r} = \vec{A} + t(\vec{B} - \vec{A})$$ where $t$ is a scalar parameter. 3. **Calculate the direction vector:** $$\vec{B} - \vec{A} = (-1 - 3, 4 - 2, 0 - 0) = (-4, 2, 0)$$ 4. **Set up the parametric equations:** $$x = 3 - 4t$$ $$y = 2 + 2t$$ $$z = 0 + 0t = 0$$ 5. **Check if point $P(1|3|0)$ satisfies these equations:** From $x$ coordinate: $$1 = 3 - 4t \implies 4t = 3 - 1 = 2 \implies t = \frac{2}{4} = \frac{1}{2}$$ 6. **Check $y$ coordinate with $t=\frac{1}{2}$:** $$y = 2 + 2 \times \frac{1}{2} = 2 + 1 = 3$$ This matches the $y$ coordinate of $P$. 7. **Check $z$ coordinate:** $$z = 0$$ Matches the $z$ coordinate of $P$. 8. **Conclusion:** Since all coordinates match for $t=\frac{1}{2}$, point $P$ lies on the line through $A$ and $B$. **Final answer:** $$\boxed{\text{Point } P(1|3|0) \text{ lies on the line through } A \text{ and } B.}$$