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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 $s$.

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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

$$ \vec v=X-Y=(3+3\lambda+\mu,5+6\lambda-\mu,-1+\lambda-\mu) $$

This difference vector should satisfy the two equations you stated:

\begin{align*} \vec v\cdot(-1,1,1) &= 0 & \vec v\cdot(2,-1,3) &= 0 \\ -3-3\lambda-\mu+5+6\lambda-\mu-1+\lambda-\mu &= 0 & 6+6\lambda+2\mu-5-6\lambda+\mu-3+3\lambda-3\mu &= 0 \\ 1+4\lambda-3\mu &= 0 & -2+3\lambda &= 0 \\ 1 + \frac83 - 3\mu &= 0 & \frac23 &= \lambda \\ \frac{11}{9} &= \mu & \frac{14}{9}(4,5,-1) &= \vec v \\ \left(-\frac29, \frac{11}9, \frac{20}9\right) &= Y & \left(6, 9, \frac23\right) &= X \end{align*}

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.

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