# Find the vector equation of the line parallel to the plane $\pi$, perpendicular to the line $AB$ and that intercepts $s$

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 $s$.

You can start by connecting $Y=A+\mu(B-A)=(1-\mu,\mu,1+\mu)$ which is a generic point on $AB$ to $X=(4+3\lambda,5+6\lambda,\lambda)$ on $s$. The difference vector is
$$\vec v=X-Y=(3+3\lambda+\mu,5+6\lambda-\mu,-1+\lambda-\mu)$$
You might choose either of these points, i.e. $X$ or $Y$, as the starting point. Of course, any other point on that line would work just as well.