I ever asked a question to find the constraints which ensure 4 lines in 2D space form a convex quadrilateral; see link and it has been solved by @YNK author perfectly.
Now I hope to extend it and came across a new challenge. We learned that the equation of a plane not passing through the origin in 3D space can be $\{ax + by + cz - 1 = 0: a,b,c \in \mathbb{R} \land a^2+b^2+c^2 \ne 0 \}$. A plane always cuts the whole 3D space into two half-spaces, and this equation ensures that any point in the same half-space as the origin $(0,0,0)$ always makes the inequality $ax+by+cz−1<0$ hold.
Given 6 planes not passing through the origin, and consider a set $S$ formed by these planes: $$S := \{(x,y,z) \in \mathbb{R}^3: \bigwedge_{m=1}^6 a_mx + b_my +c_mz - 1 \le 0 \}.$$ My question is, which constraints on the coefficients $a_m, b_m, c_m, m=1,...,6$ of these 6 planes can guarantee that the set $S$ is a convex box (Of course the origin is in $S$ because $a_m0 + b_m0 +c_m0 - 1 =-1 < 0$. Note that in some cases, 6 planes do form a box but do not contain the origin, and therefore the box is not the set $S$ that we want.).
I think we first need to ensure that there are 8 intersection points serving as 8 vertices of the box, each point met by 3 planes. And then,....