This is an RSA with n being the product of 3 consecutive primes. We can approximate the prime taking cube root of n, and recover all the primes close to it.

```python=
from gmpy2 import iroot
from Crypto.Util.number import *
n = 842955733372614455917139215149786367998989408483882136463558684397050826784554405473281404986074268847665022114356876291445365131548244366599468837778869392604574977016160648076231535588793673536845344975989305317758463762079207183948812779114263906518115672167636134526515103825946273073248648502935673944006264386299102933514541941431848389105755893385245141801018951632390644713514409554482089598460289338073545880196262116013551058638687812839058426467147481
c = 178911853582925091074953906180040707693867299041184394859091151823053279374040732087994928027427055516599491017237986483811850621047816908739709787556523375563298051776181108938835938016314409519090707332840179868647993861754529706055629293586699860402875976104999357565613123187491160887851461728095157125430360026065237722011165355477193299183616800218141392989153891385907169749132280406008273086286095580797819646823785791242488773862940145882627972745288524
pp = iroot(n, 3)[0]

primes = []
for i in range(-10000, 10000):
p = pp + i
if n % p == 0:
primes.append(p)

assert len(primes) == 3
p,q,r = primes
phi = (p-1)*(q-1)*(r-1)
e = 65537; d = pow(e, -1, phi)

print(long_to_bytes(pow(c, d, n)))
```