The system $x+2y+3z=4$,$4x+5y+3z=5$,$3x+4y+3z=\lambda$ is consistent and $3\lambda=n+100$,then $n=$

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

Let $A$ be a $3 \times 3$ real matrix such that $A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$,$A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$,and $A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$. If $X = (x_1, x_2, x_3)^T$ and $I$ is an identity matrix of order $3$,then the system $(A - 2I)X = \begin{bmatrix} 4 \\ 1 \\ 1 \end{bmatrix}$ has:

For how many different values of $a$ does the following system of linear equations have at least two distinct solutions?
$ax + y = 0$
$x + (a + 10)y = 0$

The product of all real values of $b$ such that there is no solution to the system of equations $2x + 5y + z = 19$,$-4x + by + 6z = -42$,and $-3y - bz = 81$ is: