I need to find the equation of a plane, im given the following information:
It contains the point $3,1,2$
It is parallel to the skewing lines:
$$ Line 1 = 2i+4j+3k + \lambda (5i -j-k)$$ $$ Line 2 = 6i+2j-k + \mu (i +j+k)$$
What i find strange is that when I cross product their D.V. the resulting line is not correct as I obtain (-6j+6k) instead of (i+2j-3k)
The result should be $x +2y-3z=-1$
Edit: like a certain commentor said there is a problem with the question
I let direction vectors of line be $\vec{a}, \vec{b}$, then normal vector of plane is given as $\vec{a}\times \vec{b} = \langle 0, -6 , 6 \rangle$ so the plane equattion is $-6y+6z = d$ where we find $d = 6$ by satisfying the point $(3,1,2)$. So the equation is $y-z+1=0$
See you can check whether the given answer is correct or not. Since $\langle1,3,2\rangle \cdot \langle 5,-1,-1\rangle = 0$ but $\langle1,3,2\rangle \cdot \langle 1,1,1\rangle \neq 0$ we conclude that given answer is wrong as normal vector of plane is not normal to direction vector of one of the line.