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