Line Intersecting Three Lines

 
Proposition. Let $l_1,l_2,l_3$ be lines in $P(ℝ^4)$ that do not intersect pairwise. Then there are an infinite number of lines in $l$ in $P(ℝ^4)$ that intersect $l_1,l_2,l_3$. Proof. Let $l_j=P(U_j),j=1,2,3$. Then $U_1,U_2,U_3$ are vector subspaces of $ℝ^4$, $\dim U_j=2$. Since $l_1\cap l_2=\emptyset$, $U_1\cap U_2=\{0\}$. Hence $ℝ^4=U_1\oplus U_2$, counting dimensions. Let $v\in U_3$ be non-zero, then $v=u_1+u_2,u_1\in U_1,u_2\in U_2$. If $u_1=0$ then $[v]\in l_2\cap l_3$❌ If $u_2=0$ then $[v]\in l_1\cap l_3$❌ Let $U_4=\langle u_1,u_2\rangle\leℝ^4$. Then $\dim U_4=2$. So $l_4=P(U_4)$ is a line in $P(ℝ^4)$. Also, $l_4$ contains $[u_1]\in l_1$ and $[u_2]\in l_2$ and $[u_1+u_2]=[v]\in l_3$. So $l_4$ intersects $l_1,l_2,l_3$. Unique such line intersecting $l_3$ in the point $[v]$. By construction, there exists such a line through each point $[v]$ in $l_3$, so there are infinitely many such lines $l$. β–‘ Here is a more difficult proof. Desargues’ theorem Let $P,Q_1,Q_2,Q_3,R_1,R_2,R_3$ be distinct points in a projective plane $P(V)$, such that the lines $Q_1R_1,Q_2R_2,Q_3R_3$ are distinct and concurrent at $P$. Let $S_1$ be the intersection of $Q_2Q_3,R_2R_3$ Let $S_2$ be the intersection of $Q_3Q_1,R_3R_1$ Let $S_3$ be the intersection of $Q_1Q_2,R_1R_2$ Then $S_1,S_2,S_3$ are collinear. Proof. Choose representatives $p,q_i,r_i$ such that $p=q_i+r_i$. As $p=q_2+r_2=q_3+r_3$, put $s_1:=q_2-q_3=r_3-r_2$. Since $Q_2,Q_3$ are distinct, $s_1$ is nonzero. Hence $s_1$ represents a point on the line $Q_2Q_3$. Similarly, $s_1$ represents a point on the line $R_2R_3$. Hence $s_1$ represents $S_1$, the intersection of $Q_2Q_3,R_2R_3$. Similarly, put $s_2:=q_3-q_1=r_1-r_3,s_3:=q_1-q_2=r_2-r_1$, and then $s_2,s_3$ represent $S_2,S_3$, Now, $s_1+s_2+s_3=(q_2-q_3)+(q_3-q_1)+(q_1-q_2)=0$. Hence $S_1,S_2,S_3$ are collinear, as $S_1,S_2,S_3$ are linearly dependent. β–‘