Let $S$ be the set of all column matrices $\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$ such that $b_1, b_2, b_3 \in \mathbb{R}$ and the system of equations (in real variables) $-x + 2y + 5z = b_1$; $2x - 4y + 3z = b_2$; $x - 2y + 2z = b_3$ has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each $\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} \in S$?

$A=\begin{bmatrix}-1&2&5\\2&-4&3\\1&-2&2\end{bmatrix},\quad b=\begin{bmatrix}b_{1}\\b_{2}\\b_{3}\end{bmatrix}$ $S=\{b:\ Ax=b \text{ has a solution}\}$ $\text{For which of (1)--(4) does } Bx=b \text{ have a solution } \forall b\in S?$ $\text{i.e. } \operatorname{Col}(A)\subseteq \operatorname{Col}(B)$ $1.\ \text{find } S: \text{ consistency of } Ax=b$ $2.\ Bx=b \text{ solvable } \forall b\in S \iff \operatorname{Col}(A)\subseteq\operatorname{Col}(B)$ $3.\ \text{test each option}$ $Ax=b \text{ solvable} \iff b\in \operatorname{Col}(A)=\big(\operatorname{Nul}(A^{T})\big)^{\perp}$ $\text{find } \ell \text{ with } \ell^{T}A=0$ $R_{1}+7R_{2}=(13,-26,26)=13R_{3}$ $\Rightarrow R_{1}+7R_{2}-13R_{3}=0 \Rightarrow \ell=(1,7,-13)$ $\ell\cdot b=0:\ b_{1}+7b_{2}-13b_{3}=0$ $S=\{b:\ b_{1}+7b_{2}-13b_{3}=0\}\ (\text{a plane through } O)$ $Bx=b \text{ solvable } \forall b\in S \iff \operatorname{Col}(A)\subseteq\operatorname{Col}(B)$ $\iff \operatorname{Nul}(B^{T})\subseteq \operatorname{span}\{(1,7,-13)\}$ $B_{1}=\begin{bmatrix}1&2&3\\0&4&5\\1&2&6\end{bmatrix}, \quad \det B_{1}=12\neq 0$ $\therefore \operatorname{Col}(B_{1})=\mathbb{R}^{3}\supseteq \operatorname{Col}(A) \ \checkmark$ $B_{2}=\begin{bmatrix}1&1&3\\5&2&6\\-2&-1&-3\end{bmatrix},\ \det B_{2}=0,\ \operatorname{Nul}(B_{2}^{T})=\operatorname{span}\{(1,1,3)\}$ $b=(7,-1,0)\in S \text{ but } b_{1}+b_{2}+3b_{3}=6\neq 0 \Rightarrow \text{no solution} \ \times$ $B_{3}=\begin{bmatrix}-1&2&-5\\2&-4&10\\1&-2&5\end{bmatrix}:\ R_{2}=-2R_{1},\ R_{3}=-R_{1}$ $\operatorname{rank}(B_{3})=1 \Rightarrow \operatorname{Col}(B_{3}) \nsupseteq \operatorname{Col}(A) \ \times$ $B_{4}=\begin{bmatrix}6&2&5\\2&0&3\\1&4&-5\end{bmatrix}, \quad \det B_{4}=-6\neq 0$ $\therefore \operatorname{Col}(B_{4})=\mathbb{R}^{3}\supseteq \operatorname{Col}(A) \ \checkmark$ $\boxed{(1)\ \text{and}\ (4)}$