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Question 238 - PT0-002 discussion

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A penetration tester successfully performed an exploit on a host and was able to hop from VLAN 100 to VLAN 200. VLAN 200 contains servers that perform financial transactions, and the penetration tester now wants the local interface of the attacker machine to have a static ARP entry in the local cache. The attacker machine has the following:

IP Address: 192.168.1.63

Physical Address: 60-36-dd-a6-c5-33

Which of the following commands would the penetration tester MOST likely use in order to establish a static ARP entry successfully?

A.
tcpdump -i eth01 arp and arp[6:2] == 2
Answers
A.
tcpdump -i eth01 arp and arp[6:2] == 2
B.
arp -s 192.168.1.63 60-36-DD-A6-C5-33
Answers
B.
arp -s 192.168.1.63 60-36-DD-A6-C5-33
C.
ipconfig /all findstr /v 00-00-00 | findstr Physical
Answers
C.
ipconfig /all findstr /v 00-00-00 | findstr Physical
D.
route add 192.168.1.63 mask 255.255.255.255.0 192.168.1.1
Answers
D.
route add 192.168.1.63 mask 255.255.255.255.0 192.168.1.1
Suggested answer: B

Explanation:

The arp command is used to manipulate or display the Address Resolution Protocol (ARP) cache, which is a table that maps IP addresses to physical addresses (MAC addresses) on a network. The -s option is used to add a static ARP entry to the cache, which means that it will not expire or be overwritten by dynamic ARP entries. The syntax for adding a static ARP entry is arp -s <IP address> <physical address>. Therefore, the command arp -s 192.168.1.63 60-36-DD-A6-C5-33 would add a static ARP entry for the IP address 192.168.1.63 and the physical address 60-36-DD-A6-C5-33 to the local cache of the attacker machine. This would allow the attacker machine to communicate with the target machine without relying on ARP requests or replies. The other commands are not valid or useful for establishing a static ARP entry.

asked 02/10/2024
Ravindra MG
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