服务发现
┌──(root💀kali)-[~/tryhackme/biohazard]
└─# nmap -sV -Pn 10.10.38.183
Host discovery disabled (-Pn). All addresses will be marked 'up' and scan times will be slower.
Starting Nmap 7.91 ( https://nmap.org ) at 2021-10-19 04:07 EDT
Nmap scan report for 10.10.38.183
Host is up (0.30s latency).
Not shown: 997 closed ports
PORT STATE SERVICE VERSION
21/tcp open ftp vsftpd 3.0.3
22/tcp open ssh OpenSSH 7.6p1 Ubuntu 4ubuntu0.3 (Ubuntu Linux; protocol 2.0)
80/tcp open http Apache httpd 2.4.29 ((Ubuntu))
Service Info: OSs: Unix, Linux; CPE: cpe:/o:linux:linux_kernel
Service detection performed. Please report any incorrect results at https://nmap.org/submit/ .
Nmap done: 1 IP address (1 host up) scanned in 49.55 seconds
问题:How many open ports?
答案:3
打开80端口首页,根据文字描述得知team name
问题:What is the team name in operation
答案:STARS alpha team
页面:/mansionmain/
源代码显示一行注释
It is in the /diningRoom/
打开上面提示的页面:/diningRoom/
,写了一行加密的注释:
SG93IGFib3V0IHRoZSAvdGVhUm9vbS8=
base64解密后为:
How about the /teaRoom/
点击yes导航到:/diningRoom/emblem.php
拿到emblem flag
问题:What is the emblem flag
答案:emblem{fec832623ea498e20bf4fe1821d58727}
打开/teaRoom/
,点击Lockpick.
拿到pick flag
问题:What is the lock pick flag
答案:lock_pick{037b35e2ff90916a9abf99129c8e1837}
根据提示,打开:/artRoom/
,点击YES,拿到一个目录map
Look like a map
Location:
/diningRoom/
/teaRoom/
/artRoom/
/barRoom/
/diningRoom2F/
/tigerStatusRoom/
/galleryRoom/
/studyRoom/
/armorRoom/
/attic/
在/barRoom/
填入lock_pick{037b35e2ff90916a9abf99129c8e1837}
跳转到/barRoom357162e3db904857963e6e0b64b96ba7/
点击read
,显示
Look like a music note
NV2XG2LDL5ZWQZLFOR5TGNRSMQ3TEZDFMFTDMNLGGVRGIYZWGNSGCZLDMU3GCMLGGY3TMZL5
上面密文是base_32,解密出来是:
music_sheet{362d72deaf65f5bdc63daece6a1f676e}
问题:What is the music sheet flag
答案:music_sheet{362d72deaf65f5bdc63daece6a1f676e}
输入到flag,来到/barRoom357162e3db904857963e6e0b64b96ba7/barRoomHidden.php
点击Yes,拿到gold emblem
gold_emblem{58a8c41a9d08b8a4e38d02a4d7ff4843}
Look like you can put something on the emblem slot, refresh the previous page
问题:What is the gold emblem flag
答案:gold_emblem{58a8c41a9d08b8a4e38d02a4d7ff4843}
回到/diningRoom
输入上面的flag后拿到一行密文:
klfvg ks r wimgnd biz mpuiui ulg fiemok tqod. Xii jvmc tbkg ks tempgf tyi_hvgct_jljinf_kvc
看提示上面这一段是Vigenère加密,去到这个网站解密出来是:
there is a shield key inside the dining room. The html page is called the_great_shield_key
打开/diningRoom/the_great_shield_key.html
拿到shield_key
问题:What is the shield key flag
答案:shield_key{48a7a9227cd7eb89f0a062590798cbac}
在/diningRoom2F/
的注释找到一行密文:
Lbh trg gur oyhr trz ol chfuvat gur fgnghf gb gur ybjre sybbe. Gur trz vf ba gur qvavatEbbz svefg sybbe. Ivfvg fnccuver.ugzy
经验证这个是凯撒加密,偏移量是:13
解密出来是:
You get the blue gem by pushing the status to the lower floor. The gem is on the diningRoom first floor. Visit sapphire.html
打开/diningRoom/sapphire.html
,拿到blue_jewel
问题:What is the blue gem flag
答案:blue_jewel{e1d457e96cac640f863ec7bc475d48aa}
把上面flag输入到tigerStatusRoom
,获得另一段密文:
crest 1:
S0pXRkVVS0pKQkxIVVdTWUpFM0VTUlk9
Hint 1: Crest 1 has been encoded twice
Hint 2: Crest 1 contanis 14 letters
Note: You need to collect all 4 crests, combine and decode to reavel another path
The combination should be crest 1 + crest 2 + crest 3 + crest 4. Also, the combination is a type of encoded base and you need to decode it
crest 1先base64,再base32解出来是:RlRQIHVzZXI6IG
在/galleryRoom/note.txt
找到crest 2
crest 2:
GVFWK5KHK5WTGTCILE4DKY3DNN4GQQRTM5AVCTKE
Hint 1: Crest 2 has been encoded twice
Hint 2: Crest 2 contanis 18 letters
Note: You need to collect all 4 crests, combine and decode to reavel another path
The combination should be crest 1 + crest 2 + crest 3 + crest 4. Also, the combination is a type of encoded base and you need to decode it
crest 2先base32转,再base58转得到:h1bnRlciwgRlRQIHBh
在/armorRoom/
输入shield_key后跳转页面点击reame得到crest 3
crest 3:
MDAxMTAxMTAgMDAxMTAwMTEgMDAxMDAwMDAgMDAxMTAwMTEgMDAxMTAwMTEgMDAxMDAwMDAgMDAxMTAxMDAgMDExMDAxMDAgMDAxMDAwMDAgMDAxMTAwMTEgMDAxMTAxMTAgMDAxMDAwMDAgMDAxMTAxMDAgMDAxMTEwMDEgMDAxMDAwMDAgMDAxMTAxMDAgMDAxMTEwMDAgMDAxMDAwMDAgMDAxMTAxMTAgMDExMDAwMTEgMDAxMDAwMDAgMDAxMTAxMTEgMDAxMTAxMTAgMDAxMDAwMDAgMDAxMTAxMTAgMDAxMTAxMDAgMDAxMDAwMDAgMDAxMTAxMDEgMDAxMTAxMTAgMDAxMDAwMDAgMDAxMTAwMTEgMDAxMTEwMDEgMDAxMDAwMDAgMDAxMTAxMTAgMDExMDAwMDEgMDAxMDAwMDAgMDAxMTAxMDEgMDAxMTEwMDEgMDAxMDAwMDAgMDAxMTAxMDEgMDAxMTAxMTEgMDAxMDAwMDAgMDAxMTAwMTEgMDAxMTAxMDEgMDAxMDAwMDAgMDAxMTAwMTEgMDAxMTAwMDAgMDAxMDAwMDAgMDAxMTAxMDEgMDAxMTEwMDAgMDAxMDAwMDAgMDAxMTAwMTEgMDAxMTAwMTAgMDAxMDAwMDAgMDAxMTAxMTAgMDAxMTEwMDA=
Hint 1: Crest 3 has been encoded three times
Hint 2: Crest 3 contanis 19 letters
Note: You need to collect all 4 crests, combine and decode to reavel another path
The combination should be crest 1 + crest 2 + crest 3 + crest 4. Also, the combination is a type of encoded base and you need to decode it
crest 3先base64转,得到一串二进制数字,2进制转文本得到一个16进制的串,16进制转ASCII得到:c3M6IHlvdV9jYW50X2h
在/attic/
输入输入shield_key后跳转页面点击reame得到crest 4
crest 4:
gSUERauVpvKzRpyPpuYz66JDmRTbJubaoArM6CAQsnVwte6zF9J4GGYyun3k5qM9ma4s
Hint 1: Crest 2 has been encoded twice
Hint 2: Crest 2 contanis 17 characters
Note: You need to collect all 4 crests, combine and decode to reavel another path
The combination should be crest 1 + crest 2 + crest 3 + crest 4. Also, the combination is a type of encoded base and you need to decode it
crest 4先base58解码得到一个16进制字符串,再转ASCII得到:pZGVfZm9yZXZlcg==
所以crest 1 + crest 2 + crest 3 + crest 4得到一个字符串:
RlRQIHVzZXI6IGh1bnRlciwgRlRQIHBhc3M6IHlvdV9jYW50X2hpZGVfZm9yZXZlcg==
base64解密出来是:
FTP user: hunter, FTP pass: you_cant_hide_forever
所以我们现在得到了ftp的登入权限,登录到靶机,把文件全部下载下来待分析:
┌──(root💀kali)-[~/tryhackme/biohazard]
└─# ftp 10.10.38.183
Connected to 10.10.38.183.
220 (vsFTPd 3.0.3)
Name (10.10.38.183:root): hunter
331 Please specify the password.
Password:
230 Login successful.
Remote system type is UNIX.
Using binary mode to transfer files.
ftp> ls -alh
200 PORT command successful. Consider using PASV.
150 Here comes the directory listing.
drwxrwxrwx 2 1002 1002 4096 Sep 20 2019 .
drwxrwxrwx 2 1002 1002 4096 Sep 20 2019 ..
-rw-r--r-- 1 0 0 7994 Sep 19 2019 001-key.jpg
-rw-r--r-- 1 0 0 2210 Sep 19 2019 002-key.jpg
-rw-r--r-- 1 0 0 2146 Sep 19 2019 003-key.jpg
-rw-r--r-- 1 0 0 121 Sep 19 2019 helmet_key.txt.gpg
-rw-r--r-- 1 0 0 170 Sep 20 2019 important.txt
important.txt文件透露出隐藏文件夹:
┌──(root💀kali)-[~/tryhackme/biohazard]
└─# cat important.txt 127 ⨯
Jill,
I think the helmet key is inside the text file, but I have no clue on decrypting stuff. Also, I come across a /hidden_closet/ door but it was locked.
From,
Barry
问题:Where is the hidden directory mentioned by Barry
答案:/hidden_closet/
001-key.jpg用steghide 分离出一个文件:
┌──(root💀kali)-[~/tryhackme/biohazard]
└─# steghide extract -sf 001-key.jpg
Enter passphrase:
wrote extracted data to "key-001.txt".
┌──(root💀kali)-[~/tryhackme/biohazard]
└─# cat key-001.txt
cGxhbnQ0Ml9jYW
key1字符串:cGxhbnQ0Ml9jYW
002-key.jpg的文件信息Comment里有一个奇怪的字符串,我们先记录下来:
┌──(root💀kali)-[~/tryhackme/biohazard]
└─# exiftool 002-key.jpg
ExifTool Version Number : 12.16
File Name : 002-key.jpg
Directory : .
File Size : 2.2 KiB
File Modification Date/Time : 2021:10:19 23:46:31-04:00
File Access Date/Time : 2021:10:19 23:47:26-04:00
File Inode Change Date/Time : 2021:10:19 23:46:31-04:00
File Permissions : rw-r--r--
File Type : JPEG
File Type Extension : jpg
MIME Type : image/jpeg
JFIF Version : 1.01
Resolution Unit : None
X Resolution : 1
Y Resolution : 1
Comment : 5fYmVfZGVzdHJveV9
Image Width : 100
Image Height : 80
Encoding Process : Progressive DCT, Huffman coding
Bits Per Sample : 8
Color Components : 3
Y Cb Cr Sub Sampling : YCbCr4:2:0 (2 2)
Image Size : 100x80
Megapixels : 0.008
key2字符串:5fYmVfZGVzdHJveV9
key3用binwalk分离出一个文件:
┌──(root💀kali)-[~/tryhackme/biohazard]
└─# binwalk -e 003-key.jpg
DECIMAL HEXADECIMAL DESCRIPTION
--------------------------------------------------------------------------------
0 0x0 JPEG image data, JFIF standard 1.01
1930 0x78A Zip archive data, at least v2.0 to extract, uncompressed size: 14, name: key-003.txt
2124 0x84C End of Zip archive, footer length: 22
┌──(root💀kali)-[~/tryhackme/biohazard]
└─# cd _003-key.jpg.extracted
┌──(root💀kali)-[~/tryhackme/biohazard/_003-key.jpg.extracted]
└─# ls
78A.zip key-003.txt
┌──(root💀kali)-[~/tryhackme/biohazard/_003-key.jpg.extracted]
└─# cat key-003.txt
3aXRoX3Zqb2x0
key3字符串:3aXRoX3Zqb2x0
结合key1+key2+key3得到的字符串是:
cGxhbnQ0Ml9jYW5fYmVfZGVzdHJveV93aXRoX3Zqb2x0
这个是base64串,解出来是:
plant42_can_be_destroy_with_vjolt
问题:Password for the encrypted file
答案:plant42_can_be_destroy_with_vjolt
用上面的秘钥 解密gpg文件
┌──(root💀kali)-[~/tryhackme/biohazard]
└─# gpg --decrypt helmet_key.txt.gpg > helmet_key.txt
gpg: AES256 encrypted data
gpg: 以 1 个密码加密
┌──(root💀kali)-[~/tryhackme/biohazard]
└─# cat helmet_key.txt
helmet_key{458493193501d2b94bbab2e727f8db4b}
问题:What is the helmet key flag
答案:helmet_key{458493193501d2b94bbab2e727f8db4b}
用上面的key打开隐藏页面/hidden_closet/
点击READ,得到一串密文:
wpbwbxr wpkzg pltwnhro, txrks_xfqsxrd_bvv_fy_rvmexa_ajk
去到这个网站,利用vigenere-solver
解密得到:
weasker login password, stars_members_are_my_guinea_pig
去到/studyroom
,输入helmet_key,解锁房间拿到里面的gz文件,解压后得到ssh登录用户名:
SSH user: umbrella_guest
点击EXAMINE,得到ssh密码:
SSH password: T_virus_rules
问题:What is the SSH login username
答案:umbrella_guest
问题:What is the SSH login password
答案:T_virus_rules
全局寻找STARS bravo team
:
find / |xargs grep -ri 'STARS bravo team' -l >f.txt
在/var/www/html/hiddenCloset8997e740cb7f5cece994381b9477ec38/index.php
找到文本:
umbrella_guest@umbrella_corp:/tmp$ cat /var/www/html/hiddenCloset8997e740cb7f5cece994381b9477ec38/index.php
<html>
<head>
<title>Closet room</title>
<h1 align="center">Closet room</h1>
</head>
<body>
<img alt="closet_room" src="../images/maxresdefault.jpg.5" style="display: block;margin-left: auto;margin-right: auto; width: 50%;"/>
<p>The closet room lead to an underground cave</p>
<p>In the cave, Jill met injured Enrico, the leader of the STARS Bravo team. He mentioned there is a traitor among the STARTS Alpha team.</p>
<p>When he was about to tell the traitor name, suddenly, a gun shot can be heard and Enrico was shot dead.</p>
<p>Jill somehow cannot figure out who did that. Also, Jill found a MO disk 1 and a wolf Medal</p>
<p><b>Read the MO disk 1?</b> <a href="MO_DISK1.txt">READ</a></p>
<p><b>Examine the wolf medal?</b> <a href="wolf_medal.txt">EXAMINE</a></p>
</body>
</html>
问题:Who the STARS bravo team leader
答案:Enrico
全局寻找Chris
:
find / |xargs grep -ri 'Chris' -l >c.txt
在/home/umbrella_guest/.jailcell/chris.txt
找到文本:
umbrella_guest@umbrella_corp:/tmp$ cat /home/umbrella_guest/.jailcell/chris.txt
Jill: Chris, is that you?
Chris: Jill, you finally come. I was locked in the Jail cell for a while. It seem that weasker is behind all this.
Jil, What? Weasker? He is the traitor?
Chris: Yes, Jill. Unfortunately, he play us like a damn fiddle.
Jill: Let's get out of here first, I have contact brad for helicopter support.
Chris: Thanks Jill, here, take this MO Disk 2 with you. It look like the key to decipher something.
Jill: Alright, I will deal with him later.
Chris: see ya.
MO disk 2: albert
问题:Where you found Chris
答案:jailcell
问题:Who is the traitor
答案:weasker
albert
其实就是上面密串wpbwbxr wpkzg pltwnhro, txrks_xfqsxrd_bvv_fy_rvmexa_ajk
的维吉利亚解密密串,解出来是:
weasker login password, stars_members_are_my_guinea_pig
问题:The login password for the traitor
答案:stars_members_are_my_guinea_pig
切换到weasker
,查看sudo -l,发现这个账号拥有root的一切权限
umbrella_guest@umbrella_corp:/var/www/html$ su weasker
Password:
weasker@umbrella_corp:/var/www/html$ sudo -l
[sudo] password for weasker:
Matching Defaults entries for weasker on umbrella_corp:
env_reset, mail_badpass, secure_path=/usr/local/sbin\:/usr/local/bin\:/usr/sbin\:/usr/bin\:/sbin\:/bin\:/snap/bin
User weasker may run the following commands on umbrella_corp:
(ALL : ALL) ALL
直接sudo su切换到root,拿到root.txt
weasker@umbrella_corp:/var/www/html$ sudo su
root@umbrella_corp:/var/www/html# cat /root/root.txt
In the state of emergency, Jill, Barry and Chris are reaching the helipad and awaiting for the helicopter support.
Suddenly, the Tyrant jump out from nowhere. After a tough fight, brad, throw a rocket launcher on the helipad. Without thinking twice, Jill pick up the launcher and fire at the Tyrant.
The Tyrant shredded into pieces and the Mansion was blowed. The survivor able to escape with the helicopter and prepare for their next fight.
The End
flag: 3c5794a00dc56c35f2bf096571edf3bf
问题:The name of the ultimate form
答案:Tyrant
问题:The root flag
答案:3c5794a00dc56c35f2bf096571edf3bf
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