Consider the system of equations: $ax + by + cz = 2$, $bx + cy + az = 2$, $cx + ay + bz = 2$, where $a, b, c$ are real numbers such that $a + b + c = 0$. Then, the system

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 = $

Use the product $\left[\begin{array}{lll}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]\left[\begin{array}{lll}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$ to solve the system of equations:
$x-y+2z=1$
$2y-3z=1$
$3x-2y+4z=2$

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

The number of solutions of the following equations $x_2 - x_3 = 1$,$-x_1 + 2x_3 = -2$,$x_1 - 2x_2 = 3$ is

Consider the system of equations: $x + y - az = 1$; $2x + ay + z = 1$; $ax + y - z = 2$. Which of the following statements is correct?