Consider the following system of equations
$\begin{array}{l}\mathrm{\alpha x}+2\mathrm{y}+\mathrm{z}=1\\ 2\mathrm{\alpha x}+3\mathrm{y}+\mathrm{z}=1\\ 3\mathrm{x}+\mathrm{\alpha y}+2\mathrm{z}=\mathrm{\beta }\end{array}$
For some $\mathrm{\alpha },\mathrm{\beta }\in \mathrm{ℝ}$. Then which of the following is NOT correct.

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{ccc}\mathrm{\alpha }& 2& 1\\ 2\mathrm{\alpha }& 3& 1\\ 3& \mathrm{\alpha }& 2\end{array}\right|=\mathrm{\alpha }(6-\mathrm{\alpha })-2(4\mathrm{\alpha }-3)+1\left(2{\mathrm{\alpha }}^2-9\right)\\ ={\mathrm{\alpha }}^2-2\mathrm{\alpha }-3\\ =(\mathrm{\alpha }-3)(\mathrm{\alpha }+1)\\ {\mathrm{\Delta }}_1=\left|\begin{array}{lll}1& 2& 1\\ 1& 3& 1\\ \mathrm{\beta }& \mathrm{\alpha }& 2\end{array}\right|=1(6-\mathrm{\alpha })-2(2-\mathrm{\beta })+1(\mathrm{\alpha }-3\mathrm{\beta })\\ =2-\mathrm{\beta }\\ {\mathrm{\Delta }}_2=\left|\begin{array}{ccc}\mathrm{\alpha }& 1& 1\\ 2\mathrm{\alpha }& 1& 1\\ 3& \mathrm{\beta }& 2\end{array}\right|=\mathrm{\alpha }(2-\mathrm{\beta })-2\mathrm{\alpha }(2-\mathrm{\beta })+3\left(0\right)\\ =-2\mathrm{\alpha }+\mathrm{\alpha \beta }=\mathrm{\alpha }(-2+\mathrm{\beta })\\ {\mathrm{\Delta }}_3=\left|\begin{array}{ccc}\mathrm{\alpha }& 2& 1\\ 2\mathrm{\alpha }& 3& 1\\ 3& \mathrm{\alpha }& \mathrm{\beta }\end{array}\right|=\mathrm{\alpha }(3\mathrm{\beta }-\mathrm{\alpha })-2(2\mathrm{\alpha \beta }-3)+1\left(2{\mathrm{\alpha }}^2-9\right)\\ ={\mathrm{\alpha }}^2-\mathrm{\alpha \beta }-3\\ \mathrm{\alpha }=-1,\mathrm{\beta }\ne 2\Rightarrow \mathrm{\Delta }=0,{\mathrm{\Delta }}_1\ne 0\\ \Rightarrow \text{ no solution }\\ \mathrm{\alpha }\ne -1,\mathrm{\beta }=2\Rightarrow \mathrm{\Delta }\ne 0\\ \Rightarrow \text{ unique solution }\\ \mathrm{\alpha }=3,\mathrm{\beta }\ne 2\Rightarrow \mathrm{\Delta }=0,{\mathrm{\Delta }}_1\ne 0\\ \Rightarrow \text{ no solution }\\ \mathrm{\alpha }=-1,\mathrm{\beta }\in \mathrm{ℝ}\Rightarrow \mathrm{no} \mathrm{solution} \mathrm{or} \mathrm{infinitely} \mathrm{many} \mathrm{sol}. \end{array}$