Let $P _ { 1 }$ and $P _ { 2 }$ be two planes given by
$$\begin{aligned}
& P _ { 1 } : 10 x + 15 y + 12 z - 60 = 0 \\
& P _ { 2 } : \quad - 2 x + 5 y + 4 z - 20 = 0
\end{aligned}$$
Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P _ { 1 }$ and $P _ { 2 }$ ?\\
(A) $\frac { x - 1 } { 0 } = \frac { y - 1 } { 0 } = \frac { z - 1 } { 5 }$\\
(B) $\frac { x - 6 } { - 5 } = \frac { y } { 2 } = \frac { z } { 3 }$\\
(C) $\frac { x } { - 2 } = \frac { y - 4 } { 5 } = \frac { z } { 4 }$\\
(D) $\frac { x } { 1 } = \frac { y - 4 } { - 2 } = \frac { z } { 3 }$