For the system $S$ of linear equations $x+y+z=3, 2x+2y-z=3, x+y+\lambda z=1$, the incorrect option among the following statements is:

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 $x+4y-5z=b_3$

If the system of linear equations : $x+y+2z=6$,$2x+3y+az=a+1$,$-x-3y+bz=2b$ where $a, b \in R$,has infinitely many solutions,then $7a+3b$ is equal to :

If the system of linear equations $x + ay + z = 3$,$x + 2y + 2z = 6$,and $x + 5y + 3z = b$ has no solution,then:

Consider the system of linear equations $x+y+z=5$,$x+2y+\lambda^2 z=9$,and $x+3y+\lambda z=\mu$,where $\lambda, \mu \in R$. Then,which of the following statements is $NOT$ correct?

If the unique solution of the simultaneous linear equations $3x - 2y + z = 5k$, $2x + 3y - 2z = -5k$, and $x + 4y + 3z = k$ is $x = \alpha, y = \beta, z = 3$, then $k =$