Answer:
We have the next relation:
A = (b*d)/c
because we have direct variation with b and d, but inversely variation with c.
Now, if we have 3d instead of d, we have:
A' = (b*(3d))/c
now, we want A' = A. If b,c, and d are the same in both equations, we have that:
3bd/c = b*d/c
this will only be true if b or/and d are equal to 0.
If d remains unchanged, and we can play with the other two variables we have:
3b'd/c' = bd/c
3b'/c' = b/c
from this we can took that: if c' = c, then b' = b/3, and if b = b', then c' = 3c.
Of course, there are other infinitely large possible combinations that are also a solution for this problem where neither b' = b or c' = c
