Let $P_{1} {and P}_{2}$ be two planes given by
$P_{1} : 10 x + 15 y + 12 z - 60 = 0,$
$P_{2} : - 2 x + 5 y + 4 z - 20 = 0$
Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_{1} {and P}_{2}$ ?

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$\frac{x}{1} = \frac{y - 4}{- 2} = \frac{z}{3}$

$\frac{x - 6}{- 5} = \frac{y}{2} = \frac{z}{3}$

$\frac{x - 1}{0} = \frac{y - 1}{0} = \frac{z - 1}{5}$

$\frac{x}{- 2} = \frac{y - 4}{5} = \frac{z}{4}$

answer is A, B, D.

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Detailed Solution

The line should be either coincident on $P_{1}$ or on $P_{2}$ or interest on $P_{1}$ and $P_{2}$ on different points.
$D) \frac{x}{1} = \frac{y - 4}{- 2} = \frac{z}{3} = λ$

$\Rightarrow (λ, - 2 λ + 4, 3 λ)$ lie on $P_{2}$
$A) \frac{x - 1}{0} = \frac{y - 1}{0} = \frac{z - 1}{5}$ intersects $P_{1}$ and $P_{2}$ on different points.
$B) \frac{x - 6}{5} = \frac{y}{2} = \frac{z}{3}$ also intersects $P_{1} a n d P_{2}$ on different points.