Given matrix, and it's eigen value is 1,
I write $|A-1.I|=$\begin{vmatrix} 0& -1&2 \\ 0&0 &0 \\ 1& 2& 0\end{vmatrix}$$
Now I have to find an Eigen-vector(obviously non-zero) such that $|A-I|.X=0$
and this is $\begin{vmatrix} 0 & -1 & 2\\ 0 & 0 &0 \\ 1 & 2 &0 \end{vmatrix}$.$\begin{bmatrix} a\\ b\\ c \end{bmatrix}=0$
Now, you should think what combinations of columns 1,2 and 3 should I take to get the zero vector $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$
And, you can agree if I say, -4times the first column +2 times the second column and 1 times the third column will give you the zero vector
i.e.
$-4\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}+$ $2\begin{bmatrix} -1\\ 0\\ 2 \end{bmatrix}+$$\begin{bmatrix} 2\\ 0\\ 0 \end{bmatrix}=$$\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$
so our X becomes X=$\begin{bmatrix} -4\\ 2\\ 1 \end{bmatrix}$
And any linear combination of this X would also result in $|A-I|cX=c\{|A-I|.X\}=c.0=0$ provided $c \not=0$ otherwise we would get the zero vector and the zero vector is a trivial solution to AX=0.
Hence, answer (B)