We need the lines which can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$.

1. Key geometric idea

If two faces of a tetrahedron lie on planes $P_1$ and $P_2$, then these two faces intersect along an edge of the tetrahedron.

So the required line must be the line of intersection of the two planes $P_1$ and $P_2$.

Thus, among the given options, we must find which lines coincide with $P_1 \cap P_2.$

2. Find the intersection line of $P_1$ and $P_2$

The planes are: $P_1: 10x+15y+12z-60=0$ $P_2: -2x+5y+4z-20=0$

So, $10x+15y+12z=60 \quad ...(1)$ $-2x+5y+4z=20 \quad ...(2)$

Multiply (2) by $5$: $-10x+25y+20z=100$

Add to (1): $40y+32z=160$ $5y+4z=20$ $y=\frac{20-4z}{5} \quad ...(3)$

Substitute in (2): $-2x+5\left(\frac{20-4z}{5}\right)+4z=20$ $-2x+20-4z+4z=20$ $-2x=0$ $x=0.$

Hence the intersection line is given by $x=0, \qquad 5y+4z=20.$

Let $z=t$. Then $y=\frac{20-4t}{5}=4-\frac{4t}{5}.$

A convenient parametric form is obtained by taking $t=5\lambda$: $x=0, \qquad y=4-4\lambda, \qquad z=5\lambda.$

So the line is $(x,y,z)=(0,4,0)+\lambda(0,-4,5).$

Any equivalent direction vector is acceptable, e.g. $(0,4,-5)$.

3. Check each option

Option A

$\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{5}$ This means $x=1,\quad y=1,\quad z=1+5t.$ So direction vector is $(0,0,5)$.

This line is parallel to the $z$-axis and does not satisfy $x=0$ or $5y+4z=20$ for all points. Hence it is not the intersection line.

So A is incorrect.

Option B

$\frac{x-6}{-5}=\frac{y}{2}=\frac{z}{3}=t$ So $x=6-5t,\quad y=2t,\quad z=3t.$

Check if every point lies on both planes.

For $P_1$: $10x+15y+12z-60=10(6-5t)+15(2t)+12(3t)-60$ $=60-50t+30t+36t-60=16t.$ This is not identically zero.

Hence line is not in $P_1$; so it cannot be the common edge. Thus B is incorrect.

Option C

$\frac{x}{-2}=\frac{y-4}{5}=\frac{z}{4}=t$ So $x=-2t,\quad y=4+5t,\quad z=4t.$

For $P_1$: $10x+15y+12z-60=10(-2t)+15(4+5t)+12(4t)-60$ $=-20t+60+75t+48t-60=103t,$ not identically zero.

So C is incorrect.

Option D

$\frac{x}{1}=\frac{y-4}{-2}=\frac{z}{3}=t$ So $x=t,\quad y=4-2t,\quad z=3t.$

For $P_1$: $10x+15y+12z-60=10t+15(4-2t)+12(3t)-60$ $=10t+60-30t+36t-60=16t,$ not identically zero.

So D is incorrect.

4. Conclusion

None of the given lines equals the line of intersection of $P_1$ and $P_2$.

Therefore, none of the options can be the edge common to the two faces lying on $P_1$ and $P_2$.

So my derived answer is: $\boxed{\text{None of A, B, C, D}}$

5. Comparison with stored answer

Stored correct answer: A, B, D

This does not match the geometric requirement. The common edge of the two faces must lie on both planes simultaneously, i.e. it must be their line of intersection. None of A, B, C, D satisfy both plane equations identically.

Hence I disagree with the stored answer.