Edge of Tetrahedron Formed by Two Given Planes

To determine which of the given lines can be an edge of a tetrahedron whose two faces lie on the planes $P_1$ and $P_2$, let us analyze the geometric configuration.

Let the four vertices of the tetrahedron be $A, B, C, D$.
Suppose the face $\triangle ABC$ lies on the plane $P_1$ and the face $\triangle ABD$ lies on the plane $P_2$.
Then:

• The common edge $AB$ must lie on the line of intersection $L = P_1 \cap P_2$.
• The edges $AC$ and $BC$ lie entirely in plane $P_1$ and intersect the line $L$ at $A$ and $B$, respectively.
• The edges $AD$ and $BD$ lie entirely in plane $P_2$ and intersect the line $L$ at $A$ and $B$, respectively.
• The edge $CD$ connects a point $C \in P_1 \setminus L$ and a point $D \in P_2 \setminus L$.

Thus, a straight line $\mathcal{L}$ can contain an edge of such a tetrahedron if and only if:

1. $\mathcal{L}$ lies entirely in $P_1$ (or $P_2$) and intersects the line $L = P_1 \cap P_2$ at some vertex, OR
2. $\mathcal{L}$ intersects $P_1$ and $P_2$ at two distinct points $C$ and $D$ with $C \neq D$ (i.e., neither point lies on the line of intersection $L$).

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

Let us examine each option:

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Option (A):

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

Any point on this line can be written as $(1, 1, z)$.

• Intersection with $P_1$:
  $10(1) + 15(1) + 12z - 60 = 0 \implies 12z = 35 \implies z = \frac{35}{12}$
  Thus, point $C = \left(1, 1, \frac{35}{12}\right) \in P_1$.

• Intersection with $P_2$:
  $-2(1) + 5(1) + 4z - 20 = 0 \implies 4z = 17 \implies z = \frac{17}{4}$
  Thus, point $D = \left(1, 1, \frac{17}{4}\right) \in P_2$.

Since $C \neq D$, the segment $CD$ can form the edge connecting vertex $C \in P_1$ and vertex $D \in P_2$. Hence, this line can be an edge.

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Option (B):

$$\frac{x-6}{-5} = \frac{y}{2} = \frac{z}{3}$$

Any point on this line can be written as $(6 - 5t, 2t, 3t)$.

• Intersection with $P_1$:
  $10(6-5t) + 15(2t) + 12(3t) - 60 = 0 \implies 16t = 0 \implies t = 0$
  Thus, point $C = (6, 0, 0) \in P_1$.

• Intersection with $P_2$:
  $-2(6-5t) + 5(2t) + 4(3t) - 20 = 0 \implies 32t - 32 = 0 \implies t = 1$
  Thus, point $D = (1, 2, 3) \in P_2$.

Since $C \neq D$, the segment $CD$ can form the edge connecting vertex $C \in P_1$ and vertex $D \in P_2$. Hence, this line can be an edge.

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Option (C):

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

Any point on this line can be written as $(-2t, 4 + 5t, 4t)$.

• Intersection with $P_1$:
  $10(-2t) + 15(4+5t) + 12(4t) - 60 = 0 \implies 103t = 0 \implies t = 0$
  Thus, point $C = (0, 4, 0)$.

• Intersection with $P_2$:
  $-2(-2t) + 5(4+5t) + 4(4t) - 20 = 0 \implies 45t = 0 \implies t = 0$
  Thus, point $D = (0, 4, 0)$.

Since $C = D = (0, 4, 0)$, this line does not lie in either plane and intersects both planes at the exact same single point on the line of intersection $L$. Thus, it contains only one vertex of the tetrahedron and cannot be an edge.

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Option (D):

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

Any point on this line can be written as $(t, 4 - 2t, 3t)$.

• Substituting into $P_2$:
  $-2(t) + 5(4-2t) + 4(3t) - 20 = (-2 - 10 + 12)t + (20 - 20) = 0$
  This holds for all $t \in \mathbb{R}$, which means the line lies entirely in plane $P_2$.

• Substituting into $P_1$:
  $10(t) + 15(4-2t) + 12(3t) - 60 = 16t = 0 \implies t = 0$
  The line intersects $P_1$ (and therefore the line of intersection $L = P_1 \cap P_2$) at the single point $A = (0, 4, 0)$.

Since the line lies entirely in face $P_2$ and passes through a vertex $A \in L$, we can choose vertex $D$ anywhere else on this line ($D \neq A$), making $AD$ an edge of the tetrahedron lying in $P_2$. Hence, this line can be an edge.

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Conclusion:

The correct options are A, B, and D.