Let S be the set of all column matrices  $\left[\begin{array}{c}{\text{b}}_{\text{1}}\\ {\text{b}}_{\text{2}}\\ {\text{b}}_{\text{3}}\end{array}\right]$ such that  ${\text{b}}_{\text{1}}{\text{,b}}_{\text{2}}{\text{,b}}_{\text{3}}\in ℝ$  and the system of equations (in real variables)  $-{\text{\hspace{0.17em}x\hspace{0.17em}+\hspace{0.17em}2y\hspace{0.17em}+\hspace{0.17em}5z\hspace{0.17em}=\hspace{0.17em}b}}_{\text{1}},\text{\hspace{0.17em}\hspace{0.17em}2x\hspace{0.17em}}-{\text{\hspace{0.17em}4y\hspace{0.17em}+\hspace{0.17em}3z\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}b}}_2,x-2y+2z=b_3$ has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each  $\left[\begin{array}{c}{\text{b}}_{\text{1}}\\ {\text{b}}_{\text{2}}\\ {\text{b}}_{\text{3}}\end{array}\right]\in S$? 

$\Delta =\left|\begin{array}{ccc}-1& 2& 5\\ 2& -4& 3\\ 1& -2& 2\end{array}\right|=0$ given equation represent family of lines
$\Rightarrow 2x-4y+3z-b_2+\lambda (-x+2y+5z-b_1)=0$  is same as  $x-2y+2z-b_3=0$ for same $\lambda$
 $$\begin{gathered} \frac{2-\lambda }{1}=\frac{2\lambda -4}{-2}=\frac{5\lambda +3}{2}=\frac{-(b_1\lambda +b_2)}{-b_3} \\ \lambda =\frac{1}{7}and13b_3=b_1+7b_2 \end{gathered}$$
 $A\to \left|\begin{array}{ccc}1& 2& 3\\ 0& 4& 5\\ 1& 2& 6\end{array}\right|\ne 0$
$B\to \left|\begin{array}{ccc}1& 1& 3\\ 5& 2& 6\\ 2& 1& 3\end{array}\right|\ne 0=0$ as planes are non parallel is must represent family of planes for solution to exit
 $(5\mu +1)x+(2\mu +1)y+(6\mu +3)z=b_1+\mu b_2$
Must be  $2x+b+3z+b_3=0$ which is not true always
$C\to x-2y+5z=-b_1,x-2y+5z=\frac{b_2}{2},x-2y+5z=b_3$ as these are parallel for specific case 
when $-b_1=\frac{b_2}{2}=b_3$
$D\to \left|\begin{array}{ccc}1& 2& 5\\ 2& 0& 3\\ 1& 4& -5\end{array}\right|\ne 0$