I had tried a new method hope this is less intuitive and more calculative.

In all valid codewords, the hamming distance between any two valid codewords remains constant.

Example say if we have a set of 16 valid codewords, then between any two valid codewords in this set of 16, the hamming distance would be same.

The given codewords are

**0101011( Let's Say A), 1001101 (let's say B) and 1110001 ( say C)**

For finding the hamming distance between any two codewords, perform XOR operation of both and find the number of 1's present in the result.

A⊕B = 1100110 (Hamming distance is 4)

B⊕C=0111100 (Hamming distance is 4)

A⊕C=1011010(Hamming distance is 4)

So, in our given code system, all codewords are at a hamming distance of 4.

Any new valid codeword must also have a hamming distance of 4 from these 3(A, B and C) valid codewords.

Now considering each option one by one

**(I) 0010111 (Let's call it D)**

A⊕ D=111100 (Hamming distance =4)

B⊕D=1011010 (Hamming distance =4)

C⊕D = 1100110 (Hamming distance =4)

This new codeword (D) has same hamming distance with the present codewords. **Hence, this is a valid codeword.**

**(II)0110110 (Assume is as E)**

A⊕E=0011101 (Hamming distance = 4)

B⊕E = 1111011 (Hamming distance = 6)

Now, Stop here as this codeword E has a hamming distance of 6 with codeword B. But it should have same hamming distance with all the valid codewords.

**So, Choice (II) is surely Invalid.**

**(III) 1011010 (Let's call it F)**

A⊕F = 1110001 (Hamming distance =4)

B⊕F= 0010111 (Hamming distance =4)

C⊕F=0101011 (Hamming distance =4)

**Yes, this is a valid codeword.**

**(IV)0111010 (Let's call it G)**

A⊕G=0010011 (Hamming distance =3)

This is an invalid codeword.

**So, clearly, our answer is option A**