The system of linear equations $x+y+z=6, x+2y+3z=10$ and $x+2y+az=b$ has no solutions when

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

Let $S_1$ and $S_2$ be respectively the sets of all $a \in R - \{0\}$ for which the system of linear equations
$a x + 2 a y - 3 a z = 1$
$(2 a + 1) x + (2 a + 3) y + (a + 1) z = 2$
$(3 a + 5) x + (a + 5) y + (a + 2) z = 3$
has a unique solution and infinitely many solutions,respectively. Then:

Let $\alpha_1, \alpha_2$ be two values of $\alpha$ for which the system $2\alpha x + y = 5$,$x - 6y = \alpha$,and $x + y = 2$ is consistent. Then $|2(\alpha_1 + \alpha_2)|$ is -

The system of equations $2x + 6y = -11$,$6x + 20y - 6z = -3$ and $6y - 18z = -1$ are

Let $A = \begin{bmatrix} 2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b \end{bmatrix}$. If $A^3 = 4A^2 - A - 21I$,where $I$ is the identity matrix of order $3 \times 3$,then $2a + 3b$ is equal to: