Oh my, why am I being kept in the museum for so long?

As suggested in the tags, this is all about number bases. So, we have

$(CBA)_k - (ABC)_k = (12-10)k^2 + (11-11)k + (10 - 12)k^0$

$(CBA)_k - (ABC)_k = 2k^2 - 2$

Then, we have

$2k^2-2 = (776)_{\frac{k}{2}}$

$2k^2-2 = 7(\frac{k^2}{4})+7(\frac{k}{2})+6$

$8k^2-8 = 7k^2 + 14k + 24$

$k^2-14k -32 = 0$

$(k-16)(k+2) = 0$

Therefore, $k$ could be either $16$ or $-2$ . Since a base of a number is positive, so $k = \boxed{16}$ .