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

<p>$$\begin{aligned} & X^T A X=0 \\ & (x y z)\left(\begin{array}{lll} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=0 \\ & (x y z)\left(\begin{array}{l} a_1 x+a_2 y+a_3 z \\ b_1 x+b_2 y+b_3 z \\ c_1 x+c_2 y+c_3 z \end{array}\right)=0 \end{aligned}$$</p> <p>$$\begin{aligned} & x\left(a_1 x+a_2 y+a_3 z\right)+y\left(b_1 x+b_2 y+b_3 z\right) \\ & +z\left(c_1 x+c_2 y+c_3 z\right)=0 \\ & a_1=0, b_2=0 c_3=0 \\ & a_2+b_1=0, a_3+c_1=0, b_3=c_2=0 \end{aligned}$$</p> <p>$$\begin{aligned} & A=\text { skew symm matrix } \\ & A=\left(\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right) ; \quad A=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)=\left(\begin{array}{l} 1 \\ 4 \\ -5 \end{array}\right) \\ & \Rightarrow A=\left(\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right)\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)=\left(\begin{array}{l} 1 \\ 4 \\ -5 \end{array}\right) \end{aligned}$$</p> <p>$$\begin{aligned} & x+y=1 \\ & -x+z=4 \\ & y+z=5 \\ & \left(\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right)\left(\begin{array}{l} 1 \\ 2 \\ 1 \end{array}\right)=\left(\begin{array}{l} 1 \\ 4 \\ -8 \end{array}\right) \end{aligned}$$</p> <p>$$\begin{array}{ll} 2 x+y=0 & x=-1 \\ -x+z=4 & y=2 \\ -y-2 z=-8 & z=3 \end{array}$$</p> <p>$$\begin{aligned} & A=\left(\begin{array}{ccc} 0 & -1 & 2 \\ 1 & 0 & 3 \\ -2 & -3 & 0 \end{array}\right) \\ & 2(\mathrm{~A}+\mathrm{I})=\left(\begin{array}{ccc} 2 & -2 & 4 \\ 2 & 2 & 6 \\ -2 & -6 & 2 \end{array}\right) \\ & 2(\mathrm{~A}+\mathrm{I})=120 \Rightarrow \operatorname{det}|\operatorname{adi}(2(\mathrm{~A}+\mathrm{I}))| \\ & =120^2=2^6 \cdot 3^2 \cdot 5^2 \\ & \alpha=6, \beta=2, \gamma=2 \end{aligned}$$</p>