Let $A = \begin{bmatrix} 1 & -4 & 7 \\ 0 & 3 & -5 \\ -2 & 5 & -9 \end{bmatrix}$ and $B = \begin{bmatrix} a \\ -b \\ -c \end{bmatrix}$. If $A$ and $[A: B]$ have the same rank, then:

Consider the system of equations in $x, y$ and $z$:
$12x + by + cz = 0$
$ax + 24y + cz = 0$
$ax + by + 36z = 0$
(where $a, b, c$ are real numbers,$a \ne 12, b \ne 24, c \ne 36$).
If the system of equations has a non-trivial solution $(z \ne 0)$,then the value of $\frac{1}{a - 12} + \frac{2}{b - 24} + \frac{3}{c - 36}$ is:

If the system of equations $x + y - z = 0, 3x - \alpha y - 3z = 0, x - 3y + z = 0$ has a non-zero solution,then $\alpha = $

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

If the system of linear equations $2x - 3y = \gamma + 5$ and $\alpha x + 5y = \beta + 1$,where $\alpha, \beta, \gamma \in R$,has infinitely many solutions,then the value of $|9\alpha + 3\beta + 5\gamma|$ is equal to

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