Let P _ 1 and P _ 2 be two planes given by P _ 1 : 10x + 15y + 12z...

Given,

${P}_{1} and${P}_{2}$be two planes given by</p><p>${P}_{1}: 10x + 15y + 12z - 60 = 0,$</p><p>${P}_{2}: - 2x + 5y + 4z - 20 = 0$.</p><p>Now finding line of intersection of both the planes,</p><p>Let$z = \lambda$then we have equations,</p><p>$10x + 15y = 60 - 12\lambda . . . \left(1\right)$</p><p>$- 2x + 5y = 20 - 4\lambda . . . . . \left(2\right)$</p><p>Now on solving equation \left(1\right) \text{ and } \left(2\right)$ we get,

$\dfrac{x}{0} = \dfrac{y - 4}{- 4} = \dfrac{z}{5}$ \left\{\text{Replacing}\lambda\text{with}z\right\}$</p><p>Now any skew line with the line of intersection of given plane can be edge of tetrahedron,</p><p>Or any line which is intersecting with the line of intersection passing through plane${P}_{1}$or plane${P}_{2}$can be edge of tetrahedron, </p><p>Now using above concept we will solve all options,</p><p>Now form option$\left(A\right)$we have,</p><p>$\dfrac{x - 1}{0} = \dfrac{y - 1}{0} = \dfrac{z - 1}{5}$, any point lying on this line will be \left(1, 1, 5\lambda + 1\right)$ now satisfying this point in given planes we have,

$10 \times 1 + 15 \times 1 + 12 \times \left(5\lambda + 1\right) - 60 = 0$

$⇒ 60\lambda = 23 ⇒ \lambda = \dfrac{23}{60}$ now we can see line is intersecting the plane at some point,

Now checking for plane $\left(2\right)$ we have,

$- 2 \times 1 + 5 \times 1 + 4 \times \left(5\lambda + 1\right) - 20 = 0$

$⇒ 20\lambda = 13 ⇒ \lambda = \dfrac{13}{20}$

So, we can say line is also intersecting plane ${P}_{2}$ for some value of $\lambda$, So this line can be edge of tetrahedron.

Now solving for option \left(B\right)\dfrac{x - 6}{- 5} = \dfrac{y}{2} = \dfrac{z}{3}$any point on the line will be \left(- 5\lambda + 6, 2\lambda, 3\lambda\right)$

Similarly putting in plane ${P}_{1} \text{ and } {P}_{2}$ we get value of $\lambda$ as ${\lambda}_{{P}_{1}} = 0 \text{ and } {\lambda}_{{P}_{2}} = 1$

So, it can be the edge of tetrahedron.

Now solving for option \left(C\right)\dfrac{x}{- 2} = \dfrac{y - 4}{5} = \dfrac{z}{4}$any point on this line will be \left(- 2\lambda, 5\lambda + 4, 4\lambda\right)$

Similarly checking in plane we get,

${\lambda}_{{P}_{1}} = 0 \text{ and } {\lambda}_{{P}_{2}} = 0$, so it is passing through one point and not lying on plane

So, it cannot be the edge of tetrahedron.

Now checking for option $\left(D\right)\dfrac{x}{1} = \dfrac{y - 4}{- 2} = \dfrac{z}{3}$ now any point on this line will be \left(\lambda, - 2\lambda + 4, 3\lambda\right)$and for$\lambda = 0$point will be \left(0, - 4, 0\right)$ which is lying on line of intersection and DR of plane ${P}_{2}$ is \left(- 2, 5, 4\right)$and DR of line is \left(1, - 2, 3\right)$,

Now line is lying completely on plane ${P}_{2}$ as $- 2 \times 1 + 5 \times \left(- 2\right) + 3 \times 4 = 0$

Hence, it can be the edge of tetrahedron.