Let the given system be written as

$A\mathbf{x}=\mathbf{b},$ where

We are given that

$S=\left\{\mathbf{b}\in \mathbb R^3: A\mathbf{x}=\mathbf{b}\text{ has at least one solution}\right\}.$

We must find which option systems have a solution for every $\mathbf{b}\in S$.

1. First find the set $S$

A vector $\mathbf{b}$ belongs to $S$ iff it lies in the column space of $A$. So we first determine the consistency condition for the given system.

Row relation in $A$

Observe the rows of $A$:

$R_1=(-1,2,5),\quad R_2=(2,-4,3),\quad R_3=(1,-2,2).$

Now,

$R_1+R_2=(-1+2,\,2-4,\,5+3)=(1,-2,8),$ which is not equal to $R_3$.

Let us compute rank by elimination:

Take $R_3\to R_3+R_1$:

$R_3+R_1=(0,0,7\mid b_1+b_3).$

Take $R_2\to R_2+2R_1$:

$R_2+2R_1=(0,0,13\mid b_2+2b_1).$

So,

For consistency, the last two equations in $z$ must agree:

Thus

$\frac{b_2+2b_1}{13}=\frac{b_1+b_3}{7}.$

Cross-multiplying:

$7(b_2+2b_1)=13(b_1+b_3)$ $7b_2+14b_1=13b_1+13b_3$ $b_1+7b_2-13b_3=0.$

Hence

$S=\left\{\begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix}: b_1+7b_2-13b_3=0\right\}.$

So $S$ is a plane in $\mathbb R^3$.

2. Key idea for each option

If an option has coefficient matrix $M$, then the system

$M\mathbf{x}=\mathbf{b}$

has a solution for every $\mathbf{b}\in S$ iff

$S\subseteq \operatorname{Col}(M).$

Since $S$ is a 2-dimensional plane, this means the column space of $M$ must contain exactly that plane (or all of $\mathbb R^3$ if rank $3$).

Equivalently, if $\operatorname{rank}(M)=2$, then its consistency condition must be the same plane equation:

$b_1+7b_2-13b_3=0.$

We now test each option.

3. Option A

System:

Coefficient matrix:

Compute determinant:

More directly by expansion along first row:

So $M_A$ is invertible. Therefore for every $\mathbf{b}\in\mathbb R^3$, the system has a unique solution. Hence certainly for every $\mathbf{b}\in S$.

So A is correct.

4. Option B

System:

Coefficient matrix:

Find the consistency condition by finding a linear relation among rows. Suppose

$\alpha R_1+\beta R_2+\gamma R_3=0.$

This gives

From coordinates:

$\alpha+5\beta-2\gamma=0$ $\alpha+2\beta-\gamma=0$ $3\alpha+6\beta-3\gamma=0 \iff \alpha+2\beta-\gamma=0.$

So only first two are independent. Subtracting,

$3\beta-\gamma=0 \implies \gamma=3\beta,$ then

$\alpha+2\beta-3\beta=0 \implies \alpha=\beta.$

Thus one row relation is

$R_1+R_2+3R_3=0.$

Hence consistency requires the same relation on RHS:

$b_1+b_2+3b_3=0.$

But vectors in $S$ satisfy

$b_1+7b_2-13b_3=0,$ which is a different plane. So not every $\mathbf{b}\in S$ will satisfy option B.

For example, take $\mathbf{b}=(13,0,1)^T$. Then

$13+7(0)-13(1)=0,$ so $\mathbf{b}\in S$. But

$13+0+3\neq 0,$ so option B is inconsistent.

Therefore B is incorrect.

5. Option C

System:

Coefficient matrix:

Observe:

$R_2=-2R_1, \qquad R_3=-R_1.$

So all rows are multiples of one row. Hence rank is $1$. Thus consistency requires

This is a line in $\mathbb R^3$, not the whole plane $S$. Therefore it certainly cannot hold for every $\mathbf{b}\in S$.

For example, take again $\mathbf{b}=(13,0,1)^T\in S$. But here we would need

$b_2=-2b_1=-26,$ which fails.

Hence C is incorrect.

6. Option D

System:

Coefficient matrix:

Compute determinant:

Now,

egin{vmatrix}0&3\\4&-5\end{vmatrix}=0(-5)-12=-12, egin{vmatrix}2&3\\1&-5\end{vmatrix}=-10-3=-13, egin{vmatrix}2&0\\1&4\end{vmatrix}=8.

So,

Thus $M_D$ is invertible, so the system has a unique solution for every $\mathbf{b}\in\mathbb R^3$, hence for every $\mathbf{b}\in S$.

So D is correct.

7. Final conclusion

The systems that have at least one solution for each $\mathbf{b}\in S$ are:

$\boxed{A, D}.$

This matches the stored correct answer.