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 \(\displaystyle l_1(x,y,z)=(1,0,1)+t(2,-1,-2)\) and \(\displaystyle l_2(x,y,z)=(2,-5,0)+s(-1,2,1)\) intercept on point A also interpect on plane \(\displaystyle \pi:-x+2y-z+4=0\) in point B and C, decide area of triangle ABC

Progress:
Point A:
If we equal them we get \(\displaystyle t=\frac{7}{3}\) and \(\displaystyle s=-\frac{11}{3}\) that means \(\displaystyle A=(\frac{17}{3},-\frac{7}{3},-\frac{8}{3})\)
Point B:
We replace \(\displaystyle x,y,z\) in the plane with \(\displaystyle l_1\) and get \(\displaystyle t=1\)
so \(\displaystyle B=(3,-1,-1)\)
Point C:
We replace \(\displaystyle x,y,z\) in the plane with \(\displaystyle l_2\) and get \(\displaystyle s=2\) that means \(\displaystyle C=(0,-1,2)\)
Remember area of a triangle is half of the area of paralellogram that means
\(\displaystyle \text{Area}=\frac{1}{2}\left|\textbf{AB}\times \textbf{AC} \right|\)
and \(\displaystyle AB=(-\frac{8}{3},\frac{4}{3},-\frac{5}{3})\)
\(\displaystyle AC=(-\frac{17}{3},\frac{4}{3},\frac{14}{3})\)
so the area is \(\displaystyle \frac{\sqrt{45881}}{19}\) 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 \(\displaystyle (\frac{76}{9}, \frac{197}{9}, 4)\)

Regards,
\(\displaystyle |\pi\rangle\)

 

[MarkFL]

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

[MATH]\text{Area}=\frac{1}{2}\left|\textbf{AB}\times \textbf{AC} \right|[/MATH]

to get:

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