How to get a 12 bytes authentication tag in AES-GCM

How to get a 12 bytes authentication tag in AES-GCM

i want to generate a 12 bytes authentication in AES-GCM. my code only give me an output tag of 16 bytes. i have tried to use the min_tag_length but it rises an error. can i get your guidance, thank in advance.

import os
from cryptography.hazmat.primitives.ciphers import (
    Cipher, algorithms, modes
)

#f(StoC)= SC || IC || GMAC( SC || AK || StoC)
#associated data , AAD= SC || AK || StoC
SC=bytes.fromhex("10")
StoC=bytes.fromhex("39373635313934303931363137383231")
key = bytes.fromhex("000102030405060708090A0B0C0D0E0F") #EK
A_key = bytes.fromhex("000102030405060708090A0B0C0D0E0F") #AK
#AAD
associated_data=SC+A_key+StoC
#print (associated_data.hex(),type(associated_data))


def encrypt(key, plaintext, associated_data):
    #global associated_data
    #global key

    # Generate a random 96-bit IV.
    part1 = bytes.fromhex("485845033225519F") #Syst_title
    part2 = bytes.fromhex("00000002") # IC

    # Concatenate the parts
    iv = part1 + part2

    #iv = bytes.fromhex(sum)
    #print (type(iv))

    # Construct an AES-GCM Cipher object with the given key and a
    encryptor = Cipher(
        algorithms.AES(key),
        modes.GCM(iv, tag_length=12),
    ).encryptor()

    # associated_data will be authenticated but not encrypted,
    # it must also be passed in on decryption.
    encryptor.authenticate_additional_data(associated_data)

    # Encrypt the plaintext and get the associated ciphertext.
    # GCM does not require padding.
    ciphertext = encryptor.update(plaintext) + encryptor.finalize()
    iv_hex=iv.hex()
    ciphertext_hex = ciphertext.hex()
    encryptor.tag_hex= encryptor.tag.hex()
    print(f"{iv_hex=}")
    #print(f"{ciphertext_hex=}")
    print(f"{encryptor.tag_hex=}")

    challenge =SC+part2+encryptor.tag
    challenge_hex=challenge.hex()
    print(f"{challenge_hex=}")

    #return (iv, ciphertext, encryptor.tag)
    return challenge_hex

encrypt(key, StoC, associated_data)