The number of solutions of the following system of linear homogeneous equations $x-y+z=0$,$x+2y-z=0$,and $2x+y+3z=0$ is

Consider the system of equations:
$x - 2y + 3z = -1$; $-x + y - 2z = k$; $x - 3y + 4z = 1$
$STATEMENT-1$: The system of equations has no solution for $k \neq 3$.
$STATEMENT-2$: The determinant $\left|\begin{array}{ccc}1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4\end{array}\right| = 0$.

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 $\alpha^2+\beta^2+\gamma^2=$

If the solution of the system of simultaneous equations $\frac{1}{x}+\frac{2}{y}-\frac{3}{z}-1=0$,$\frac{2}{x}-\frac{4}{y}+\frac{3}{z}-1=0$ and $\frac{3}{x}+\frac{6}{y}-\frac{6}{z}-4=0$ is $x=\alpha, y=\beta, z=\gamma$,then $\alpha^2+\gamma^2=$

If the system of equations $x + y + z = 5$,$x + 2y + 3z = 9$,and $x + 3y + \alpha z = \beta$ has infinitely many solutions,then $\beta - \alpha$ equals:

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