$\mathrm{P}_1$ 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.