MHB Area of Triangle ABC: Calculating with Points A,B,C

[Petrus]

Hello MHB,
I am working with an old exam that I don't get same answer.
Line $$l_1(x,y,z)=(1,0,1)+t(2,-1,-2)$$ and $$l_2(x,y,z)=(2,-5,0)+s(-1,2,1)$$ intercept on point A also interpect on plane $$\pi:-x+2y-z+4=0$$ in point B and C, decide area of triangle ABC

Progress:
Point A:
If we equal them we get $$t=\frac{7}{3}$$ and $$s=-\frac{11}{3}$$ that means $$A=(\frac{17}{3},-\frac{7}{3},-\frac{8}{3})$$
Point B:
We replace $$x,y,z$$ in the plane with $$l_1$$ and get $$t=1$$
so $$B=(3,-1,-1)$$
Point C:
We replace $$x,y,z$$ in the plane with $$l_2$$ and get $$s=2$$ that means $$C=(0,-1,2)$$
Remember area of a triangle is half of the area of paralellogram that means
$$\text{Area}=\frac{1}{2}\left|\textbf{AB}\times \textbf{AC} \right|$$
and $$AB=(-\frac{8}{3},\frac{4}{3},-\frac{5}{3})$$
$$AC=(-\frac{17}{3},\frac{4}{3},\frac{14}{3})$$
so the area is $$\frac{\sqrt{45881}}{19}$$ which does not agree with the facit, what I am doing wrong?
edit: I forgot to say that cross product of AB x AC is $$(\frac{76}{9}, \frac{197}{9}, 4)$$

Regards,
$$|\pi\rangle$$

 

[MarkFL]

To denote the area with vectors and their cross, I would use the $\LaTeX$ code:

$$\text{Area}=\frac{1}{2}\left|\textbf{AB}\times \textbf{AC} \right|$$

to get:

$$\text{Area}=\frac{1}{2}\left|\textbf{AB}\times \textbf{AC} \right|$$