If the lines x/2 = (y+1)/3 = z/4 and x/3 = (y-k)/2 = z/1 intersect, then k is equal to:

The point on the line x/2 = (y+1)/3 = z/4 = a is (2a, 3a-1, 4a).
The point on the line x/3 = (y-k)/2 = z/1 = b is (3b, 2b+k, b).
Now, the given condition is lines are intersecting, we get
2a = 3b
3a - 1 = 2b + k
4a = b
From 2a = 3b and 4a = b, we have 2a = 3(4a), which implies 2a = 12a, so a = 0.
Then b = 4a = 0.
Substituting a = 0 and b = 0 in 3a - 1 = 2b + k, we get
3(0) - 1 = 2(0) + k
-1 = k
k = -1
Hence, option 'C' is correct.