There are two equations of line given:
$x - y + z = 0 = 2x - 3y + 4z$
$x + y + 2z - 3 = 0 = 2x + 3y + 3z - 4$.
We need to find the length and equation of shortest distance between these lines. I got the length as $13/\sqrt{66}$ using projection method, but I am not getting the second part of finding the equation.
The answer is given as $3x - y - z = 0 = x + 2y + z - 1$. Please help with the steps/hints involved to reach here.
The two line given are respectively $l_1$ and $l_2$; $P_1=0, Q_1=(1,2,1)\in l_1; P_2=(5,-2,0), Q_2=(2,-1,1) \in l_2$ $$\vec{P_1Q_1}=(1,2,1), \vec{P_2Q_2}=(-3,1,1), \vec{n}:=\vec{P_1Q_1}\times \vec{P_2Q_2}=(1,-4,7)$$ $l_1=\mathbb R (1,2,1), l_2=(5,-2,0)+\mathbb R(-3,1,1)$
We are looking for an equation of the plane containing $l_1$ and directed by $\vec{n}$. A vector orthogonal to this plane is $\vec{n}\times\vec{P_1Q_1}=-6(3,-1,-1)$. So $(x,y,z)\in \mathbb R^3$ is in this plane iff $3x-y-z=0$.
We are looking for an equation of the plane containing $l_2$ and directed by $\vec{n}$. A vector orthogonal to this plane is $\vec{n}\times\vec{P_2Q_2}=-11(1,2,1)$. So $M=(x,y,z)\in \mathbb R^3$ is in this plane $\iff \langle\vec{Q_2M},(1,2,1)\rangle=0 \iff x+2y+z-1=0$
Hence the answer given.