Let be a 3 × 3 matrix such that XT AX = O for all non zero $3\times 1\text{ matrices }\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\text{. }$
If $\mathrm{A}\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{l} 1\\ 4\\ -5\end{array}\right],\mathrm{A}\left[\begin{array}{l}1\\ 2\\ 1\end{array}\right]=\left[\begin{array}{l} 0\\ 4\\ -8\end{array}\right]$, and $\det \left(\mathrm{adj}\right(2(\mathrm{A}+\mathrm{I})\left)\right)=2^{\mathrm{\alpha }}{3}^{\mathrm{\beta }}{5}^{\mathrm{\gamma }},\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma },\in \mathrm{N}$, then ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2$ is 

$\begin{array}{l}{\mathrm{X}}^{\mathrm{T}}\mathrm{AX}=0\\ \left(\mathrm{xyz}\right)\left(\begin{array}{lll}{\mathrm{a}}_1& {\mathrm{a}}_2& {\mathrm{a}}_3\\ {\mathrm{b}}_1& {\mathrm{b}}_2& {\mathrm{b}}_3\\ {\mathrm{c}}_1& {\mathrm{c}}_2& {\mathrm{c}}_3\end{array}\right)\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right)=0\\ \left(\mathrm{xyz}\right)\left(\begin{array}{c}{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2\mathrm{y}+{\mathrm{a}}_3\mathrm{z}\\ {\mathrm{b}}_1\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{b}}_3\mathrm{z}\\ {\mathrm{c}}_1\mathrm{x}+{\mathrm{c}}_2\mathrm{y}+{\mathrm{c}}_3\mathrm{z}\end{array}\right)=0\end{array}$
$\begin{array}{l}\mathrm{x}\left({\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2\mathrm{y}+{\mathrm{a}}_3\mathrm{z}\right)+\mathrm{y}\left({\mathrm{b}}_1\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{b}}_3\mathrm{z}\right)\\ +\mathrm{z}\left({\mathrm{c}}_1\mathrm{x}+{\mathrm{c}}_2\mathrm{y}+{\mathrm{c}}_3\mathrm{z}\right)=0\\ {\mathrm{a}}_1=0,{\mathrm{b}}_2=0{\mathrm{c}}_3=0\\ {\mathrm{a}}_2+{\mathrm{b}}_1=0,{\mathrm{a}}_3+{\mathrm{c}}_1=0,{\mathrm{b}}_3={\mathrm{c}}_2=0\end{array}$
A = skew symm matrix
$\begin{array}{l}\mathrm{A}=\left(\begin{array}{ccc}0& \mathrm{x}& \mathrm{y}\\ -\mathrm{x}& 0& \mathrm{z}\\ -\mathrm{y}& -\mathrm{z}& 0\end{array}\right);\mathrm{A}=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)=\left(\begin{array}{c}1\\ 4\\ -5\end{array}\right)\\ \Rightarrow \mathrm{A}=\left(\begin{array}{ccc}0& \mathrm{x}& \mathrm{y}\\ -\mathrm{x}& 0& \mathrm{z}\\ -\mathrm{y}& -\mathrm{z}& 0\end{array}\right)\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)=\left(\begin{array}{c}1\\ 4\\ -5\end{array}\right)\end{array}$
$\begin{array}{l}\mathrm{x}+\mathrm{y}=1\\ -\mathrm{x}+\mathrm{z}=4\\ \mathrm{y}+\mathrm{z}=5\\ \left(\begin{array}{ccc}0& \mathrm{x}& \mathrm{y}\\ -\mathrm{x}& 0& \mathrm{z}\\ -\mathrm{y}& -\mathrm{z}& 0\end{array}\right)\left(\begin{array}{c}1\\ 2\\ 1\end{array}\right)=\left(\begin{array}{c}1\\ 4\\ -8\end{array}\right)\\ 2\mathrm{x}+\mathrm{y}=0\begin{array}{c}\mathrm{x}=-1\\ -\mathrm{x}+\mathrm{z}=4\end{array}\mathrm{y}=2\\ -\mathrm{y}-2\mathrm{z}=-8\\ \mathrm{z}=3\\ \mathrm{A}=\left(\begin{array}{ccc}0& -1& 2\\ 1& 0& 3\\ -2& -3& 0\end{array}\right)\\ 2(\mathrm{A}+\mathrm{I})=\left(\begin{array}{ccc}2& -2& 4\\ 2& 2& 6\\ -2& -6& 2\end{array}\right)\end{array}$
$\begin{array}{l}2(\mathrm{A}+\mathrm{I})=120\Rightarrow \det \left|\mathrm{adi}\right(2(\mathrm{A}+\mathrm{I})\left)\right|\\ ={120}^2=2^6\cdot 3^2\cdot 5^2\\ \mathrm{\alpha }=6,\mathrm{\beta }=2,\mathrm{\gamma }=2\end{array}$