1. **State the problem:** Find the point where the line passing through points $(4,-4,9)$ and $(1,2,3)$ intersects the $zx$-plane. 2. **Formula and explanation:** The parametric form of a line through points $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ is: $$x = x_1 + t(x_2 - x_1), \quad y = y_1 + t(y_2 - y_1), \quad z = z_1 + t(z_2 - z_1)$$ where $t$ is a parameter. 3. **Apply to given points:** $$x = 4 + t(1 - 4) = 4 - 3t$$ $$y = -4 + t(2 + 4) = -4 + 6t$$ $$z = 9 + t(3 - 9) = 9 - 6t$$ 4. **Condition for $zx$-plane:** The $zx$-plane is defined by $y=0$. Set $y=0$ to find $t$: $$-4 + 6t = 0 \implies 6t = 4 \implies t = \frac{2}{3}$$ 5. **Find intersection point:** Substitute $t=\frac{2}{3}$ into $x$ and $z$: $$x = 4 - 3 \times \frac{2}{3} = 4 - 2 = 2$$ $$z = 9 - 6 \times \frac{2}{3} = 9 - 4 = 5$$ 6. **Final answer:** The line meets the $zx$-plane at the point $(2,0,5)$.