Finding the point of intersection of straight through the points A (3,-5,2) and B (11,-3,6) with the straight (r): (x,y,z) = (5,-3,2) + t (4,-2,4).
Answer:
R (7,-4,4)
$\displaystyle \vec{s} = \vec{s}_{0} + \lambda \vec{v}$
First find the direction vector of your line, i.e. $\displaystyle \vec{v} = \vec{AB} = (8, 2, 4)$.
Now, $\displaystyle s_{0}$ can simply the vector from the origin to the point A (or B, doesn't matter as long as it's on the line). So, our line is: $\displaystyle \vec{s} = (3, -5, 2) + \lambda(8,2,4)$
Now to see where lines s and r intersect, we set the them equal to each other, i.e. $\displaystyle (3, -5, 2) + \lambda(8,2,4) = (5,-3,2) + t (4,-2,4)$.