Let $A = \{X = (x, y, z)^{T} : PX = 0 \text{ and } x^{2} + y^{2} + z^{2} = 1\}$ where $P = \begin{bmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{bmatrix}$,then the set $A$:

The system of equations $\begin{cases} \lambda x+y+3 z=0 \\ 2 x+\mu y-z=0 \\ 5 x+7 y+z=0 \end{cases}$ has infinitely many solutions in $\mathbb{R}$. Then,

The system of linear equations $x + y + z = 6$,$2x + 5y + az = 36$,and $x + 2y + 3z = b$ 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 $x+4y-5z=b_3$

Let $S$ be the set of all integer solutions,$(x, y, z)$,of the system of equations
$x-2y+5z=0$
$-2x+4y+z=0$
$-7x+14y+9z=0$
such that $15 \leq x^{2}+y^{2}+z^{2} \leq 150$. Then,the number of elements in the set $S$ is equal to

If the system of linear equations: $x + y + z = 6, x + 2y + 5z = 10, 2x + 3y + \lambda z = \mu$ has infinitely many solutions,then the value of $\lambda + \mu$ equals: