Let $P_1$ and $P_2$ be two planes given by $P_1: 10x+15y+12z-60=0$, $P_2: -2x+5y+4z-20=0$. Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$?

Let P₁ and P₂ be two planes given by P₁: 10x+15y+12z-60=0, P₂: -2x+5y+4z-20=0. Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on P₁ and P₂? $P_1:\ 10x+15y+12z-60=0$ $P_2:\ -2x+5y+4z-20=0$ $\text{Which line can be an edge of a tetrahedron}$ $\text{two of whose faces lie on } P_1 \text{ and } P_2?$ $\vec n_1 \times \vec n_2 = \begin{vmatrix} \hat i & \hat j & \hat k \\ 10 & 15 & 12 \\ -2 & 5 & 4 \end{vmatrix}$ $\vec n_1 \times \vec n_2 = (0,-64,80) \parallel (0,4,-5)$ $L_0:\ x=0,\quad 5y+4z=20$ $AB \subset L_0,\qquad C \in P_1,\qquad D \in P_2$ $\ell \subset P_1 \text{ or } P_2 \text{, meeting } L_0$ $\text{or } \ell \text{ cuts } P_1, P_2 \text{ at two distinct points}$ $(1)\ \vec d=(0,0,1):\quad x=1,\ y=1$ $P_1: z=\dfrac{35}{12},\quad P_2: z=\dfrac{17}{4}\ \Rightarrow \text{distinct}$ $(2)\ \vec d=(-5,2,3):\quad (6,0,0) \in P_1$ $P_2:(1,2,3)\ \Rightarrow \text{distinct}$ $(3)\ \vec d=(-2,5,4)=\vec n_2 \Rightarrow \ell \perp P_2$ $\ell \cap P_1 = \ell \cap P_2 = (0,4,0)\ \Rightarrow \text{reject}$ $(4)\ x=t,\ y=4-2t,\ z=3t$ $-2x+5y+4z-20=0\ \forall t \Rightarrow \ell \subset P_2$ $\boxed{\text{Answer: (1), (2), (4)}}$