Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Given,

$P_1$ and $P_2$ be two planes given by

$P_1:10x+15y+12z-60=0,$

$P_2:-2x+5y+4z-20=0$.

Now finding line of intersection of both the planes,

Let $z=\lambda$ then we have equations,

$10x+15y=60-12\lambda ...\left(1\right)$

$-2x+5y=20-4\lambda .....\left(2\right)$

Now on solving equation $\left(1\right) \& \left(2\right)$ we get,

$\frac{x}{0}=\frac{y-4}{-4}=\frac{z}{5}$ $\left\{\text{Replacing }\lambda \text{with }z\right\}$

Now any skew line with the line of intersection of given plane can be edge of tetrahedron,

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, 

Now using above concept we will solve all options,

Now form option $\left(A\right)$ we have,

$\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{5}$, any point lying on this line will be $\left(1,1,5\lambda +1\right)$ now satisfying this point in given planes we have,

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

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

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

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

$\Rightarrow 20\lambda =13\Rightarrow \lambda =\frac{13}{20}$

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

Similarly putting in plane $P_1 \& P_2$ we get value of $\lambda$ as ${\lambda }_{P_1}=0 \& {\lambda }_{P_2}=1$

So, it can be the edge of tetrahedron.

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

Similarly checking in plane we get,  

${\lambda }_{P_1}=0 \& {\lambda }_{P_2}=0$, so it is passing through one point and not lying on plane 

So, it cannot be the edge of tetrahedron.

Now checking for option $\left(D\right) \frac{x}{1}=\frac{y-4}{-2}=\frac{z}{3}$ 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.