Error detection/correction algorithm
Consider any two codewords based on $b_2b_1b_0$ and $c_2c_1c_0$. The Hamming distance between the two codewords is at least twice the Hamming distance between $b_2b_1b_0$ and $c_2c_1c_0$ (why?). Furthermore, if the Hamming distance between $b_2b_1b_0$ and $c_2c_1c_0$ is exactly $1$, then your argument shows that the Hamming distance between the codewords is $3$. We can conclude that the Hamming distance is always at least $3$.