The system of equations $x+2y+3z=6$,$x+3y+5z=9$,and $2x+5y+az=12$ has no solution when $a=$

The set of real values of $\alpha$ for which the system of linear equations
$\begin{aligned}
& x+(\sin \alpha) y+(\cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& -x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}$
has a non-trivial solution is

If the system of linear equations $x + 2ay + az = 0$,$x + 3by + bz = 0$,and $x + 4cy + cz = 0$ has a non-zero solution,then $a, b, c$:

If the system of simultaneous linear equations $x+y+z=a$,$x-y+bz=2$,and $2x+3y-z=1$ has infinitely many solutions,then $b-5a=$

Statement $-1$: The system of linear equations
$x + (\sin \alpha)y + (\cos \alpha)z = 0$
$x + (\cos \alpha)y + (\sin \alpha)z = 0$
$x - (\sin \alpha)y - (\cos \alpha)z = 0$
has a non-trivial solution for only one value of $\alpha$ lying in the interval $(0, \frac{\pi}{2})$.
Statement $-2$: The equation in $\alpha$
$\left| \begin{matrix} \cos \alpha & \sin \alpha & \cos \alpha \\ \sin \alpha & \cos \alpha & \sin \alpha \\ \cos \alpha & -\sin \alpha & -\cos \alpha \end{matrix} \right| = 0$
has only one solution lying in the interval $(0, \frac{\pi}{2})$.

For the system $S$ of linear equations $x+y+z=3, 2x+2y-z=3, x+y+\lambda z=1$, the incorrect option among the following statements is: