Equation of plane is

$\left|\begin{array}{ccc}x-1& y-0& z-1\\ 1-1& 2& 1-1\\ 1-0& 0-1& 1+2\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x-1& y& z-1\\ 0& 2& 0\\ 1& -1& 3\end{array}\right|=0$

$\Rightarrow \left(x-1\right)\left(6\right)-y\left(0\right)+\left(z-1\right)\left(-2\right)=0$

$\Rightarrow 3\left(x-1\right)-z+1=0$

$\Rightarrow 3x-z-2=0$

Since, $\overrightarrow{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ is parallel to $3x-z-2=0$, therefore normal vector of plane is perpendicular to $\overrightarrow{a}$, hence

$\Rightarrow 3\alpha -\gamma =0 ...\left(1\right)$

Now,

$\overrightarrow{a}\perp \left(\dot{i}+2\hat{j}+3\hat{k}\right)$

$\Rightarrow \alpha +2\beta +3\gamma =0 ...\left(2\right)$

And,

$\overrightarrow{a}·(\hat{i}+\hat{j}+2\hat{k})=0$

$\Rightarrow \alpha +\beta +2\gamma =2 ...\left(3\right)$

On solving $\left(1\right), \left(2\right) \& \left(3\right)$, we get

$\alpha =1, \beta =-5, \gamma =3$

So, ${\left(\alpha -\beta +\gamma \right)}^2=81$