If the system of equations $x+y+z=1$, $x+2y+4z=k$ and $x+4y+10z=k^2$ is consistent, then $k$ is equal to

Let $M = (a_{ij})$,$i, j \in \{1, 2, 3\}$,be a $3 \times 3$ matrix such that $a_{ij} = 1$ if $j+1$ is divisible by $i$,otherwise $a_{ij} = 0$. Then which of the following statements is (are) true?
$(A)$ $M$ is invertible
$(B)$ There exists a nonzero column matrix $\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$ such that $M \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} = \begin{bmatrix} -a_1 \\ -a_2 \\ -a_3 \end{bmatrix}$
$(C)$ The set $\{X \in \mathbb{R}^3 : MX = 0, X \neq 0\}$ is non-empty,where $0 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$
$(D)$ The matrix $(M - 2I)$ is invertible,where $I$ is the $3 \times 3$ identity matrix

If the solution for the system of equations $x+2y-z=3$,$3x-y+2z=1$ and $2x-2y+3z=2$ is $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2=$

$A$ trust fund has Rs. $30,000$ that must be invested in two different types of bonds. The first bond pays $5 \%$ interest per year,and the second bond pays $7 \%$ interest per year. Using matrix multiplication,determine how to divide Rs. $30,000$ among the two types of bonds if the trust fund must obtain an annual total interest of Rs. $1800$.

If the following system of linear equations
$2x + y + z = 5$
$x - y + z = 3$
$x + y + az = b$
has no solution,then :

For real numbers $\alpha$ and $\beta$,consider the following system of linear equations:
$x+y-z=2, x+2y+\alpha z=1, 2x-y+z=\beta$. If the system has infinite solutions,then $\alpha+\beta$ is equal to $.....$