$A+I=\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
$A=\left[\begin{array}{ccc}0 & 0 & -2 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$
We know, $AX = \lambda X$ here $\lambda=-1$ given.
$\left[\begin{array}{ccc}0 & 0 & -2 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]\left[\begin{array}{c} a \\ b \\ c\end{array}\right] = (-1)\left[\begin{array}{c} a \\ b \\ c\end{array}\right]$
$\left[\begin{array}{c} -2c \\ -b \\ -c\end{array}\right]=\left[\begin{array}{c} -a \\ -b \\ -c\end{array}\right]$
$\therefore a=2c, b=b, c=c$
$\color{Green} Ans: \text{B.} \left[\begin{array}{c} 2t \\ s \\ t\end{array}\right],s,t \in \mathbb{R}$