Surely this hyperbolic honeycomb is possible as a mere alternation of o3x4x3o5*a only. Accordingly its edges (within an euclidean setting) would have size x = 1 and k = sqrt[2+sqrt(2)] = 1.847759 respectively.

Incidence matrix according to Dynkin symbol

o3β4β3o5*a   (N → ∞)

both( . . . .    )   | 12N |   20   20 |  10  10   30  10 | 10  2  10
---------------------+-----+-----------+------------------+----------
sefa( o3β . .    ) & |   2 | 120N    * |   1   0    1   1 |  1  1   1  x
sefa( . s4s .    )   |   2 |    * 120N |   0   1    2   0 |  2  0   1  k
---------------------+-----+-----------+------------------+----------
      o3β . .      & |   3 |    3    0 | 40N   *    *   * |  1  1   0  x3o
both( . s4s .    )   |   4 |    0    4 |   * 30N    *   * |  2  0   0  k4o
sefa( o3β4β .    ) & |   3 |    1    2 |   *   * 120N   * |  1  0   1  ox&#k = {(t,T,T)}
sefa( o3β . o5*a ) & |   5 |    5    0 |   *   *    * 24N |  0  1   1  x5o
---------------------+-----+-----------+------------------+----------
      o3β4β .      & ♦  24 |   24   48 |   8  12   24   0 | 5N  *   *  with verf [4,t,4,T,3/2,T]
      o3β . o5*a   & ♦  12 |   60    0 |  20   0    0  12 |  * 2N   *
sefa( o3β4β3o5*a )   |  10 |   10   10 |   0   0   10   2 |  *  * 12N  xo5ox&#k (tall pap variant)

starting figure: o3x4x3o5*a