QJEEAdv22Mab7d7: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Mab7d7 Solution: Let and be two planes given by Which of the following straight…

$\mathrm{P}_1$ and $\mathrm{P}_2$ be two planes given by

$\begin{aligned}& P_1: 10 x+15 y+12 z-60=0 \\\\& P_2:-2 x+5 y+4 z-20=0\end{aligned}$

Now finding line of intersection of both the planes,

Let $z=\lambda$ , then

$\begin{aligned}10 x+15 y & =60-12 \lambda .........(1) \\\\-2 x+5 y & =20-4 \lambda .........(2)\end{aligned}$

Now solving the eq. (1) and (2) we get,

$\frac{x}{0}=\frac{y-4}{-4}=\frac{z}{5} \quad(\because \lambda=z)$

Now any skew line with the line of intersection of given plane can be edge of tertrahedron.

Now using above concept we will solve all options.

For option (A)

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

Point is $(1,1,5 \lambda+1)$

Now satisfying this point in given plane we have,

$\begin{aligned}& 10 \times 1+15 \times 1+12 \times(5 \lambda+1)-60=0 \\\\& \Rightarrow 60 \lambda=23 \\\\& \Rightarrow \lambda=\frac{23}{60}\end{aligned}$

Now we can see line is intersecting the plane $\mathrm{P}_1$ , at some point.

Now checking for plane $\left(\mathrm{P}_2\right)$

$\begin{aligned}& -2 \times 1+5 \times 1 +4(5 \lambda+1)=20 \\\\& \Rightarrow 20 \lambda =13 \\\\& \Rightarrow \lambda =\frac{13}{20}\end{aligned}$

Also intersecting plane $\left(\mathrm{P}_2\right)$

Hence, it can be the edge of tetrahedron.

For option (B)

$\frac{x-6}{-5}=\frac{y}{2}=\frac{z}{3} \text { point is }(-5 \lambda+6,2 \lambda, 3 \lambda)$

this point is sytisfying plane $P_1$

$\begin{aligned}& 10(-5 \lambda+6)+15 \times 2 \lambda+12 \times 3 \lambda=60 \\\\& \Rightarrow -50 \lambda+60+30 \lambda+36 \lambda=60 \\\\& \Rightarrow 16 \lambda =0 \\\\& \Rightarrow \lambda =0\end{aligned}$

Now checking for plane $\mathrm{P}_2$

$\begin{aligned}& -2(-5 \lambda+6)+5 \times 2 \lambda+4 \times 3 \lambda=20 \\\\& \Rightarrow +10 \lambda-12+10 \lambda+12 \lambda=20 \\\\& \Rightarrow 32 \lambda=32 \\\\& \Rightarrow \lambda =1\end{aligned}$

Hence, it can be the edge of tetrahedron.

For option (C)

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

point is $(-2 \lambda, 5 \lambda+4,4 \lambda)$

Similarly checking in plane $\mathrm{P}_1$ we get.

$\begin{aligned}& \lambda \text { for } P_1=0 \\\\& \lambda \text { for } P_2=0\end{aligned}$

Hence, it can not be the edge of tetrahedron

For option (D),

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

point $(\lambda,-2 \lambda+4,3 \lambda)$ and for $\lambda=0$ point will be $(0,-4,0)$ which is lying on line of intersection and DR of plane $\mathrm{P}_2$ is $(-2,5,4)$ and DR of line is $(1,-2,3)$

Now line is lying completely on $\mathrm{P}_2$

Hence, it can be the edge of tetrahedron.