The equation $x + 2y + 3z = 1,$ $2x + y + 3z = 2,$ $5x + 5y + 9z = 4$ have

The number of values of $k$ for which the system of linear equations, $(k + 2) x + 10y = k,\,\,kx + (k + 3)y = k – 1$  has no solution, is

If the system of equations  $x+2 y+3 z=3$  ; $4 x+3 y-4 z=4$ ; $8 x+4 y-\lambda z=9+\mu$ has infinitely many solutions, then the ordered pair $(\lambda, \mu)$ is equal to

Let $p, q, r$ be nonzero real numbers that are, respectively, the $10^{\text {th }}, 100^{\text {th }}$ and $1000^{\text {th }}$ terms of a harmonic progression. Consider the system of linear equations

$x+y+z=1$

$10 x+100 y+1000 z=0$

$q r x+p r y+p q z=0$.

The correct option 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 $b_1, b_2, b_3 \in R$ and the system of equations (in real variables)

$-x+2 y+5 z=b_1$

$2 x-4 y+3 z=b_2$

$x-2 y+2 z=b_3$

has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each$\left[\begin{array}{l}b_1 \\ b_2 \\ b_3\end{array}\right]$ $\in$ $S$ ?

$(A)$ $x+2 y+3 z=b_1, 4 y+5 z=b_2$ and $x+2 y+6 z=b_3$

$(B)$ $x+y+3 z=b_1, 5 x+2 y+6 z=b_2$ and $-2 x-y-3 z=b_3$

$(C)$ $-x+2 y-5 z=b_1, 2 x-4 y+10 z=b_2$ and $x-2 y+5 z=b_3$

$(D)$ $x+2 y+5 z=b_1, 2 x+3 z=b_2$ and $x+4 y-5 z=b_3$