Suppose we know that when the three planes $P_1$, $P_2$ and $P_3$ in $\mathbb{R}_3$ intersect in pairs, we get three lines $L_1$, $L_2$, and $L_3$ which are distinct and parallel. How can I show that the three normals to $P_1$, $P_2$ and $P_3$ all lie in one plane, using an algebraic argument?
Can anyone give me some hints with intuitive explanation?
Let me try. Denote that $$(P1)\ \ \ a_1x+b_1y+c_1z +d_1=0,$$ $$(P2)\ \ \ a_2x+b_2y+c_2z +d_2=0,$$ $$(P3)\ \ \ a_3x+b_3y+c_3z + d_3=0.$$
So, $v_i = (a_i,b_i,c_i)$ is a normal of $P_i$. We have $L_1$ is intersection of $P_2$ and $P_3$. So, $L_1$ is parallel to the cross product of $v_2$ and $v_3$. Similarly, $L_2 // v_1 \times v_3$ and $L_3 // v_1 \times v_2$.
Now, because $L_1 // L_2$, we have $v_2\times v_3 // v_1\times v_3$, it implies that $v_1$, $v_2$ and $v_3$ are in the same plane (note that $v_1$ and $v_2$ is not parallel because $P_1$ intersects $P_2$).