The values of $\lambda$ and $\mu$ such that the system of equations $x+y+z=6$,$3x+5y+5z=26$,and $x+2y+\lambda z=\mu$ has no solution are:

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 $(0, \frac{\pi}{2})$.

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 -

If the system of linear equations $x + ay + z = 3$,$x + 2y + 2z = 6$,and $x + 5y + 3z = b$ has no solution,then:

If the system of equations $x+y+2z=3$,$x+2y+3z=4$ and $x+y+cz=5$ is inconsistent,then:

If the system of equations $x + y + z = 5$,$x + 2y + 3z = 9$,$x + 3y + \lambda z = \mu$ has infinitely many solutions,then the value of $\lambda + \mu$ is: