Since, the given matrix is an upper triangular one, all eigenvalues are 0. And hence $A-\lambda I=A$
So the question asks
$\begin{bmatrix} 0 &0 &\alpha \\ 0&0 &0 \\ 0& 0& 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$
what $x_1,x_2,x_3$ are suitable?
which means $x_1$times column1+$x_2$times column2+$x_3$ times column3=Zero Vector
Look carefully,if you take any linear combination of columns 1 and 2 and don't take column 3(means $x_3=0$), the result will be a zero vector.
So, if we have our X=$\begin{bmatrix} \beta\\ \gamma\\ 0 \end{bmatrix}$ where $\beta, \gamma$ are any scalar quantity,
but $x_3$ must be necessarily zero to get zero vector.
Hence, only option (B) and (D) satisfy.