Determine the parametric equations for the line that passes through the two points A=(4, -1, 5) and B= (3, 5, -1).
At what point does this line intersect the x-y plane
As you know a line equation in $\displaystyle \mathrm{R}^{3}$ has the following form:
$\displaystyle \alpha(t):=A+tu,\qquad t\in\mathbb{R},$
where $\displaystyle A$ is a point that line passes through, and $\displaystyle u\neq(0,0,0)$ is a vector that is parallel to the line.
Returning to your problem, we have a point that the desired line passes through (we may pick $\displaystyle (4, -1, 5)$ or $\displaystyle (3, 5, -1)$), but we need the vector $\displaystyle u$.
We may let $\displaystyle u$ as $\displaystyle AB$ vector, try to see that the line will be parallel to $\displaystyle AB=B-A=(-1,6,-6)$.
Thus, the line is
$\displaystyle \alpha(t):=(4,-1,5)+t(-1,6,-6),\qquad t\in\mathbb{R}.$
To find the value that line intersects $\displaystyle xy$-plane, we have to set the third component (height) of the line to $\displaystyle 0$ and solve $\displaystyle t$.
I hope you can find it now?