If the system of equations
$x+y+az=b$
$2x+5y+2z=6$
$x+2y+3z=3$
has infinitely many solutions,then $2a+3b$ is equal to $...........$.

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 $A$ be a $2 \times 2$ matrix with $\det(A)=-1$ and $\det((A+I)(\operatorname{Adj}(A)+I))=4$. Then the sum of the diagonal elements of $A$ can be.

All the real values of $p, q$ so that the system of equations $\begin{cases} 2x + py + 6z = 8 \\ x + 2y + qz = 5 \\ x + y + 3z = 4 \end{cases}$ may have no solution are

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

If $A$ is a matrix such that $\left[\begin{array}{ll} 2 & 1 \\ 3 & 2 \end{array}\right] A \left[\begin{array}{l} 1 \\ 1 \end{array}\right] = \left[\begin{array}{l} 1 \\ 0 \end{array}\right]$,then $A$ is equal to