Example input 3D array of shape (2,2,2):
[[[ 1, 2],
[ 4, 3]],
[[ 5, 6],
[ 8, 7]]]
[[0, 0, 1], # for element 2
[0, 1, 0], # for element 4
[1, 0, 1], # for element 6
[1, 1, 0]] # for element 8
argmax
argwhere
Here's an approach using np.meshgrid
to get all the indices along the first and second axes and then stacking them alongwith the max indices from the third axis using np.column_stack

d = a.argmax(1)
m,n = a.shape[:2]
c,r = np.mgrid[:m,:n]
out = np.column_stack((c.ravel(),r.ravel(),d.ravel()))
Sample run 
In [96]: a
Out[96]:
array([[[38, 49, 15, 61, 29],
[31, 88, 45, 88, 20],
[17, 97, 58, 61, 14],
[43, 77, 56, 92, 89]],
[[48, 91, 49, 35, 58],
[53, 34, 58, 92, 52],
[20, 35, 70, 41, 81],
[60, 42, 85, 82, 41]],
[[45, 41, 32, 41, 25],
[59, 32, 90, 18, 47],
[24, 93, 29, 89, 12],
[80, 27, 12, 51, 33]]])
In [97]: out
Out[97]:
array([[0, 0, 3],
[0, 1, 1],
[0, 2, 1],
[0, 3, 3],
[1, 0, 1],
[1, 1, 3],
[1, 2, 4],
[1, 3, 2],
[2, 0, 0],
[2, 1, 2],
[2, 2, 1],
[2, 3, 0]])
Alternatively, since those indices are basically repetitions, we can use np.repeat
and np.tile
to get those indices arrays and then use np.column_stack
as before, like so 
d0 = np.arange(m).repeat(n)
d1 = np.tile(np.arange(n),m)
out = np.column_stack((d0,d1,d.ravel()))