Let $P_{1}$ and $P_{2}$ be two planes given by $P_{1} : 10 x + 15 y + 12 z - 60 = 0$ , and $P_{2} : - 2 x + 5 y + 4 z - 20 = 0$ . 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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$\frac{x}{1} = \frac{y - 4}{- 2} = \frac{z}{3}$

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

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

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

answer is A, B, D.

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Detailed Solution

Line of intersection is $\frac{x}{0} = \frac{y - 4}{- 4} = \frac{z}{5}$
(1) Any skew line with the line of intersection of given planes can be edge if tetrahedron.
(2) any intersecting line with line of intersection of given planes lie either in plane p₁ and p₂ can be edge of tetralredron.