If three planes intersect pairwise along parallel lines, then their normal vectors are coplanar 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? 
 A: 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 $P_2$).
A: The cross product of any two vectors is orthogonal to each of the two vectors.
So the cross product of any two planes' normal vectors is parallel to both planes, and therefore parallel to their intersection line $\ell$.
Since the three intersection lines are parallel, $\vec{n}_1\times\vec{n}_2$ is parallel to $\vec{n}_2\times\vec{n}_3$, and we can let $\ell$ be some line parallel to these vectors. 
It follows that each of $\vec{n}_1,\vec{n}_2,\vec{n}_3$ is orthogonal to $\ell$, and therefore they are coplanar (parallel to some plane orthogonal to $\ell$.)
