Let $$M=\left[\begin{array}{ccc}0& 1& a\\ 1& 2& 3\\ 3& b& 1\end{array}\right]$$ and $$adjM=\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right]$$ where $a,b$ are real numbers. Which of the following option is/are correct?

The correct option is C If $$M\left[\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right],$$then $\alpha -\beta +\gamma =3$
$(adjM{)}_{11}=\left|\begin{array}{cc}2& 3\\ b& 1\end{array}\right|=-1$
$\Rightarrow 2-3b=-1\Rightarrow b=1$
$(adjM{)}_{22}=\left|\begin{array}{cc}0& a\\ 3& 1\end{array}\right|=-6$
$\Rightarrow -3a=-6\Rightarrow a=2$
so, $a+b=2+1=3$

Therefore,$$M=\left[\begin{array}{ccc}0& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]$$
$detM=0+-1(1-9)+2(1-6)=-2$
$|adj\left(M^2\right)|=|M^2{|}^2=|M{|}^4=(-2{)}^4=16$

$\because M^{-1}=\frac{adjM}{detM}=\frac{adjM}{-2}$
so, $(adjM{)}^{-1}+adj{M}^{-1}$
$(M^{-1}|M|{)}^{-1}+(M^{-1}{)}^{-1}|M^{-1}|$
$\frac{(M^{-1}{)}^{-1}}{|M|}+\frac{M}{|M|}$
$\frac{M}{-2}+\frac{M}{-2}=-M$
$\therefore (adjM{)}^{-1}+adj{M}^{-1}=-M$

If $$M\left[\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$$
As, $MX=B\Rightarrow X=M^{-1}B=\frac{adjM}{|M|}.B$
$=-\frac{1}{2}\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]$

$X=\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right]=\left[\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right]$
$\therefore (\alpha ,\beta ,\gamma )=(1,-1,1)$
Hence, $(\alpha -\beta +\gamma )=3$