$\left|A\right|=6\ne 0$ ( $A$ is non sigular)
Given, $6A^{-1}=a{A}^2+bA+cI$
$\Rightarrow 6I=a{A}^3+b{A}^2+cA$
$$\begin{gathered} A^2=\left[\begin{array}{ccc}0& 1& 2\\ 0& -3& 0\\ 1& 1& 1\end{array}\right]\left[\begin{array}{ccc}0& 1& 2\\ 0& -3& 0\\ 1& 1& 1\end{array}\right] \\ =\left[\begin{array}{ccc}2& -1& 2\\ 0& 9& 0\\ 1& -1& 3\end{array}\right] \end{gathered}$$
$$\begin{gathered} A^3=A^{2}A=\left[\begin{array}{ccc}2& -1& 2\\ 0& 9& 0\\ 1& -1& 3\end{array}\right]\left[\begin{array}{ccc}0& 1& 2\\ 0& -3& 0\\ 1& 1& 1\end{array}\right] \\ =\left[\begin{array}{ccc}2& 7& 6\\ 0& -27& 0\\ 3& 7& 5\end{array}\right] \end{gathered}$$
Now, $a{A}^3+b{A}^2+cA$$=\left[\begin{array}{ccc}2a+2b& 7a-b+c& 6a+2b+2c\\ 0& -27a+9b-3c& 0\\ 3a+b+c& 7a-b+c& 5a+3b+c\end{array}\right]$
$6I=\left[\begin{array}{ccc}6& 0& 0\\ 0& 6& 0\\ 0& 0& 6\end{array}\right]$
$\because a{A}^3+b{A}^2+cA=6I$

$\Rightarrow a+b=3$,
$7a-b+c=0$,
$3a+b+c=0$,
$-9a+3b-c=2$,
$5a+3b+c=6$
On solving, we get,
$a=1,b=2,c=-5$
Hence, $a+2b+3c=1+4-15=-10$