Explanation:

Let \(N = abc\) which can be written as \(100a + 10b + c\)
and \(M = xyz\) which can also be written as \(100x + 10y + z\)

\(N - M = 100(a - x) + 10(b - y) + (c - z)\)

Since, the difference has to be least;
\((a - x)\) must be minimum
If we keep \(a\) and \(x\) as consecutive numbers, the difference will always be 100 (least)
i.e. \((a - x) = 1\)

Now, \((b - y)\) must be minimum too (possibly -ve)
Minimum value of \((b - y)\) is for \(b = 1\) and \(y = 8\)
i.e. \((b - y) = -7\)

Finally, \((c - z)\) must be minimum too (possibly -ve, excluding 1 and 8)
say \(c = 2\) and \(z = 7\)
\((c - z) = -5\)
But \((a - x)\) now will be \((6 - 3) = 3\), which is direct contradiction

So, \(c\) and \(z\) must be consecutive numbers to,
say \(c = 6\) and \(z = 7\)
\((c - z) = -1\)
Therefore, \((a - x)\) now will be \((3 - 2) = 1\), Perfect!

So, \(N - M = 100(3 - 2) + 10(1 - 8) + (6 - 7)\)
i.e. \(N - M = 100 - 70 - 1 = 29\)

Col. A: \(29\)
Col. B: \(29\)

Hence, option C