The system of equations $x + y + z = 2$,$3x - y + 2z = 6$ and $3x + y + z = -18$ has

Let $S$ be the set of all column matrices $\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right]$ 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 $\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right] \in S$?
$(A)$ $x+2y+3z=b_1, 4y+5z=b_2$ and $x+2y+6z=b_3$
$(B)$ $x+y+3z=b_1, 5x+2y+6z=b_2$ and $-2x-y-3z=b_3$
$(C)$ $-x+2y-5z=b_1, 2x-4y+10z=b_2$ and $x-2y+5z=b_3$
$(D)$ $x+2y+5z=b_1, 2x+3z=b_2$ and $x+4y-5z=b_3$

For the system of linear equations $a x+y+z=1$,$x+a y+z=1$,$x+y+a z=\beta$,which one of the following statements is $NOT$ correct?

Consider the system of linear equations:
$-x+y+2z=0$
$3x-ay+5z=1$
$2x-2y-az=7$
Let $S_{1}$ be the set of all $a \in \mathbb{R}$ for which the system is inconsistent and $S_{2}$ be the set of all $a \in \mathbb{R}$ for which the system has infinitely many solutions. If $n(S_{1})$ and $n(S_{2})$ denote the number of elements in $S_{1}$ and $S_{2}$ respectively,then:

The system $2x + 3y + z = 5$,$3x + y + 5z = 7$ and $x + 4y - 2z = 3$ has

If the system of linear equations given by $x+y+z=3$,$2x+2y-z=3$,and $x+y-z=1$ is consistent and if $(x_0, y_0, z_0)$ is a solution,then $2x_0+2y_0+z_0=$