The product of all real values of $b$ such that there is no solution to the system of equations $2x + 5y + z = 19$,$-4x + by + 6z = -42$,and $-3y - bz = 81$ is:

Let $a, b, c$ be positive real numbers. The following system of equations in $x, y, z$:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1$
$\frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
$-\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
has:

If the system of equations $2x + 9y + 5z = 8$,$2x + 3y - z = -4$,$x - 2z = -5$ has an infinite number of solutions $x = -5 + at$,$y = 2 + bt$,$z = ct$,$t \in R$,then $a$,$b$,$c$ respectively are

Examine the consistency of the system of equations: $x+y+z=1$,$2x+3y+2z=2$,and $ax+ay+2az=4$.

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 -

The system of simultaneous linear equations $x-2y+3z=4$,$3x+y-2z=7$,and $2x+3y+z=6$ has