Prove that the lines p_1…

Prove that the lines $p_1:\ \frac{x-l_1}{l}=\frac{y-m_1}{m}=\frac{z-n_1}{n},\qquad p_2:\ \frac{x-l}{l_1}=\frac{y-m}{m_1}=\frac{z-n}{n_1}$ $1^\circ$ always lie in one plane, and $2^\circ$ that this plane passes through the origin.

Step 1: Notation

Let $\vec a=\{l,m,n\}$ and $\vec b=\{l_1,m_1,n_1\}$. Then $p_1$ passes through $A=\vec b$ with direction $\vec a$, and $p_2$ passes through $B=\vec a$ with direction $\vec b$.

Step 2: $1^\circ$ Coplanarity

\[ \overrightarrow{AB}\cdot(\vec a\times\vec b)=\begin{vmatrix}l-l_1&m-m_1&n-n_1\\ l&m&n\\ l_1&m_1&n_1\end{vmatrix}. \]

Adding row 3 to row 1 does not change the determinant, and it turns row 1 into $(l,m,n)$, which equals row 2. So the determinant is $0$, and the lines are coplanar.

More directly: $s=1$ on $p_1$ and $t=1$ on $p_2$ give the same point

\[ S=\vec b+\vec a=\vec a+\vec b=(l+l_1,\ m+m_1,\ n+n_1), \] so the lines intersect at $S$. This assumes $\vec a\nparallel\vec b$; the parallel case is covered in Step 4.

Step 3: $2^\circ$ The plane contains $O$

The plane goes through $A=\vec b$ and is spanned by $\vec a$ and $\vec b$. Then

\[ \overrightarrow{AO}=-\vec b\ \Longrightarrow\ \overrightarrow{AO}\cdot(\vec a\times\vec b)=-\vec b\cdot(\vec a\times\vec b)=0 . \]

So $O$ is in the plane. In coordinates, the plane is

\[ (mn_1-nm_1)x+(nl_1-ln_1)y+(lm_1-ml_1)z=0, \] and it has no constant term. The constant term would be $-\vec b\cdot(\vec a\times\vec b)=0$. $O,\ B,\ S,\ A$ are the vertices of a parallelogram.

\[p_1\cap p_2=S(l+l_1,m+m_1,n+n_1),\qquad O\in\text{plane}(p_1,p_2)\]

Step 4: The parallel case

If $\vec b=k\vec a$, then $p_1$ goes through $k\vec a$ with direction $\vec a$, so it passes through $O$. $p_2$ also passes through $O$ (take $t=-1/k$). Both are the line $O+\lambda\vec a$, so they coincide. Every plane containing that line contains $O$, so both claims still hold.