$Ax_{1} = b$ …...eq1
$Ax_{2} = b$ …...eq2
computing eq2 & eq1 we will get
$Ax_{1} = Ax_{2} $
$A(x_{2} – x_{1}) = 0$
$Ay = 0$
where $(x_{2} – x_{1}) = y= \begin{bmatrix} 3\\ 3\\3 \end{bmatrix}$
clearly $ y \neq 0 → Ay = 0 $ has non-trival solution → columns of A are Linearly Dependent
so $ky$ is also solution for $Ax = 0$ where $k$ is a scaler
Hence options$ A,B,C,D$ all of them are solutions of $Ax= 0$.