Given, if $\left|\begin{array}{ccc}2& 2k& 1\\ 1& k-1& 1\\ 2& 1& k+1\end{array}\right|=A{k}^2+Bk+C$

Applying $R_1\to R_1-R_2$

$\left|\begin{array}{ccc}2-1& 2k-k+1& 1-1\\ 1& k-1& 1\\ 2& 1& k+1\end{array}\right|=A{k}^2+Bk+C$

$\left|\begin{array}{ccc}1& k+1& 0\\ 1& k-1& 1\\ 2& 1& k+1\end{array}\right|=A{k}^2+Bk+C$

$\left[\left(k-1\right)\left(k+1\right)-1\right]-\left(k+1\right)\left(k+1-2\right)=A{k}^2+Bk+C$

$\left[k^2-1-1\right]-\left(k+1\right)\left[k-1\right]=A{k}^2+Bk+C$

$-1=A{k}^2+Bk+C$

Comparing on both sides

$A=0, B=0 \& C=-1$

Hence, $A+B+C=0+0-1=-1$