1. **State the problem:** Find the point where the line perpendicular to $L: \vec{r} = (1,-5) + s(3,5)$ passing through $P(2,0)$ intersects the y-axis. 2. **Recall the line from part a:** The direction vector of $L$ is $\vec{d} = (3,5)$. 3. **Find the direction vector of the perpendicular line:** A vector perpendicular to $\vec{d} = (3,5)$ is $\vec{d}_\perp = (-5,3)$ because the dot product $3 \times (-5) + 5 \times 3 = -15 + 15 = 0$. 4. **Write the parametric equation of the perpendicular line through $P(2,0)$:** $$\vec{r}_\perp = (2,0) + t(-5,3)$$ which gives $$x = 2 - 5t$$ $$y = 0 + 3t = 3t$$ 5. **Find the intersection with the y-axis:** On the y-axis, $x=0$. Set $x=0$ in the parametric equation: $$0 = 2 - 5t$$ Solve for $t$: $$5t = 2$$ $$t = \frac{2}{5}$$ 6. **Find the corresponding y-coordinate:** $$y = 3t = 3 \times \frac{2}{5} = \frac{6}{5}$$ 7. **Final answer:** The line intersects the y-axis at the point $\left(0, \frac{6}{5}\right)$.