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

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

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

Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $$P_1$$ and $$P_2$$?

For a tetrahedron, every edge is the intersection of two of its four face-planes.
Suppose two of those faces lie on the given planes $$P_1$$ and $$P_2$$, where

$$P_1 : 10x + 15y + 12z - 60 = 0, \qquad P_2 : -2x + 5y + 4z - 20 = 0.$$

Any edge of the tetrahedron must satisfy at least one of the following:

• it lies completely on $$P_1$$ (type I)
• it lies completely on $$P_2$$ (type II)
• it meets $$P_1$$ and $$P_2$$ in two distinct points (type III).
(In the actual tetrahedron these two intersection points become the two vertices of the edge.)

Thus a line is acceptable if

1. it is contained in at least one of the two planes,  or
2. it intersects $$P_1$$ and $$P_2$$ at two different parameter values (two different points).

We now examine each option.

Option A :  $$\dfrac{x-1}{0}=\dfrac{y-1}{0}=\dfrac{z-1}{5}=t$$

The parametric form is $$x=1,\; y=1,\; z=1+5t.$$

Intersection with $$P_1$$:

$$10(1)+15(1)+12(1+5t)-60=0$$
$$\Rightarrow -23 + 60t = 0 \;\Longrightarrow\; t=\dfrac{23}{60}.$$

Intersection with $$P_2$$:

$$-2(1)+5(1)+4(1+5t)-20=0$$
$$\Rightarrow -13 + 20t = 0 \;\Longrightarrow\; t=\dfrac{13}{20}.$$

Because the two planes are met at two different values of $$t$$, the line supplies two distinct vertices and is therefore a type III edge. Option A is possible.

Option B :  $$\dfrac{x-6}{-5}=\dfrac{y}{2}=\dfrac{z}{3}=t$$

Parametric form: $$x = 6-5t,\; y = 2t,\; z = 3t.$$

Intersection with $$P_1$$:

$$10(6-5t)+15(2t)+12(3t)-60=0$$
$$\Rightarrow 60 + 16t - 60 = 0 \;\Longrightarrow\; t = 0.$$

Intersection with $$P_2$$:

$$-2(6-5t)+5(2t)+4(3t)-20=0$$
$$\Rightarrow -32 + 32t = 0 \;\Longrightarrow\; t = 1.$$

The two planes cut the line in two different points ( $$t=0$$ and $$t=1$$ ), so the line again furnishes two distinct vertices—type III edge. Option B is possible.

Option C :  $$\dfrac{x}{-2}=\dfrac{y-4}{5}=\dfrac{z}{4}=t$$

Parametric form: $$x = -2t,\; y = 4 + 5t,\; z = 4t.$$

Intersection with $$P_1$$:

$$10(-2t)+15(4+5t)+12(4t)-60=0$$
$$\Rightarrow 103t = 0 \;\Longrightarrow\; t = 0.$$

Intersection with $$P_2$$:

$$-2(-2t)+5(4+5t)+4(4t)-20=0$$
$$\Rightarrow 45t = 0 \;\Longrightarrow\; t = 0.$$

Both planes meet the line at the same point ( $$t=0$$ ). Hence only one vertex is obtained and the entire remaining part of the line lies outside both planes. It cannot serve as an edge of the desired tetrahedron. Option C is rejected.

Option D :  $$\dfrac{x}{1}=\dfrac{y-4}{-2}=\dfrac{z}{3}=t$$

Parametric form: $$x = t,\; y = 4 - 2t,\; z = 3t.$$

Test for containment in $$P_2$$ (type II check):

$$-2x + 5y + 4z - 20$$
$$= -2t + 5(4-2t) + 4(3t) - 20$$
$$= -2t + 20 - 10t + 12t - 20 = 0$$ for every $$t$$.

Therefore the whole line lies on $$P_2$$, giving one of its faces. Intersection with $$P_1$$ occurs when

$$10t + 15(4-2t) + 12(3t) - 60 = 0$$
$$\Rightarrow 16t = 0 \;\Longrightarrow\; t = 0,$$

so the point $$\,(0,4,0)\,$ lies on both the line and $$P_1$$. The line supplies two distinct faces $$($$one is $$P_2$$ itself, the other goes through $$P_1$$ at the point $$t=0$$$)$$, hence it is a legitimate type II edge. Option D is possible.

Summarising, the lines that can be edges of the required tetrahedron are:

Option A,  Option B,  and  Option D.

Option A  ( $$\dfrac{x-1}{0}=\dfrac{y-1}{0}=\dfrac{z-1}{5}$$ ),
Option B  ( $$\dfrac{x-6}{-5}=\dfrac{y}{2}=\dfrac{z}{3}$$ ),
Option D  ( $$\dfrac{x}{1}=\dfrac{y-4}{-2}=\dfrac{z}{3}$$ ).

Hence the correct choices are:
Option A, Option B and Option D.