I have the plane:
$$\pi:2x-y+3z-1 = 0$$
$$A = (1,0,1), B = (0,1,2)$$
And $$s: X = (4,5,0) + \lambda (3,6,1)$$
I need to find a line that is perpendicular to $AB$, parallel to the plane $\pi$ and that intercepts $s$.
What I did:
Since it must be parallel to $AB = (-1,1,1)$:
$$(a,b,c)\cdot(-1,1,1) = 0$$
And since it must be parallel to the plane with normal $n=(2,-1,3)$: $$(a,b,c)\cdot(2,-1,3) = 0$$
By choosing $a=1$ and solving the system of these two equations, we get the direction vevtor $$\vec v = (4,5,-1)$$ which is exactly as the answer of the exercise. I don't know, however, how to determine a point of this line, even if I know it intersects with $s$.