Answer:
123.119.14.64/27 - subnet1
123.119.14.96/28 - subnet2
123.119.13.128/28 - subnet 3
Explanation:
123.119.14.64/26 in binary
10000000.01110111.001110.01000000/26
2^6 = 64 addresses in total; therefore we can only assign 64 addresses from this block of IP
From the question we are told that
Subnet1 supports at least 30 interfaces,
Subnet2 supports at least 12 interfaces, and
Subnet3 supports at least 9 interfaces
Subnet 1 with 30 interfaces , this means we are going to turn on 5 bits from the host bits in the IP address (starting from the right) , In doing so we will have
10000000.01110111.001110.01000000/27
123.119.14.64/27 is the nework address 0f the first subnet
Secondly we add a 1 bit (because of the broadcast address) to the first network address;
This Subnet2 supports at least 12 interfaces,
10000000.01110111.001110.01000000/27 (from the first network)
similarly we have
10000000.01110111.001110.01100000/28
Subnet3 supports at least 9 interfaces
Adding 1 bit to the second network ; we have that
10000000.01110111.001110.10000000/28,
