If the lines x/2 = y/3 = z/(4-k) and x/(1/k) = y/4 = z/5 are coplanar, then k can have exactly one value, exactly two values, exactly three values, or any value.

Let the lines be
L1: x/2 = y/3 = z/(4-k)
L2: x/(1/k) = y/4 = z/5
The direction ratios of L1 are 2, 3, 4-k
The direction ratios of L2 are 1/k, 4, 5
Let the point of intersection of L1 and L2 be (2r, 3r, (4-k)r) and ((1/k)s, 4s, 5s)
Then 2r = (1/k)s, 3r = 4s, (4-k)r = 5s
From 3r = 4s, s = (3/4)r
Substituting in 2r = (1/k)s, we get 2r = (1/k)(3/4)r
2 = 3/(4k), 8k = 3, k = 3/8
Substituting s = (3/4)r in (4-k)r = 5s, we get
(4-k)r = 5(3/4)r
4 - k = 15/4
k = 4 - 15/4 = 1/4
Since the two values of k are different, the lines are coplanar only if k = 3/8 or k = 1/4.
If the lines are coplanar, then
| 2 3 4-k |
| 1/k 4 5 | = 0
| 0 0 0 |
2(20 - 20) - 3(5/k - 5) + (4-k)(4/k - 4) = 0
-3(5/k - 5) + (4-k)(4/k - 4) = 0
-15/k + 15 + 16/k - 16 - 4 + 4k = 0
1/k + 4k - 5 = 0
1 + 4k^2 - 5k = 0
4k^2 - 5k + 1 = 0
(4k - 1)(k - 1) = 0
k = 1/4, k = 1
Thus, k can have exactly two values.