Given two lines below and that $\vec{a},\vec{b},\vec{c}$ are non complanar , find condition so that they intersect, furthermore find intersection point.
$$\vec{r_1} = \vec{a}+k(\vec{b}\times \vec{c})\\ \vec{r_2}=\vec{b}+l(\vec{c}\times \vec{a})$$
Here $\vec{a},\vec{b} $ are position vectors of two points on lines.
Trial 1
I try to manipulate but think I went wrong, I rearranged to get:
$$\vec{a}-\vec{b} = \vec{c}\times (l\vec{a}+k\vec{b})$$
Now this is never possible because left side is always in common plane of $\vec{a},\vec{b}$ and right side is always out of it.
I am wrong, obvious, but what is my mistake.
Trial 2
The lines only intersect is they are complanar, so
$$(\vec{b}-\vec{a}) \cdot ((\vec{b}\times\vec{c})\times(\vec{c}\times\vec{a}))=0\\ (\vec{b}-\vec{a})\cdot(((\vec{b}\times\vec{c})\cdot \vec{a}) \vec{c})=0$$
instantly giving $\vec{b}\cdot\vec{c}=\vec{a}\cdot\vec{c}$, which should be the condition.. I am not concerned with this, but if it contains mistake, please point.
The condition you found in the first attempt is not wrong. It means that $a-b$ is perpendicular to $c$ and perpendicular to $(la+kb)$.
Bear in mind that $a-b$ and $(la+kb)$ are co-planar, but could be mutually perpendicular for the correct choice of $k,l$.
If you do the dot product of the equation in Trial 1 with either $a$ or $b$ you can obtain $k$ or $l$ respectively, so the intersection point can be written as $$r=a+\left(\frac{a\cdot b-b^2}{b\cdot c\times a}\right)b\times c$$ for example.