If the system of equations, where a, b, c are positive,

$\mathrm{a}\mathrm{x}+2\mathrm{b}\mathrm{y}+3\mathrm{c}\mathrm{z}=0$

$2\mathrm{b}\mathrm{x}+3\mathrm{c}\mathrm{y}+\mathrm{a}\mathrm{z}=0$

$3\mathrm{c}\mathrm{x}+\mathrm{a}\mathrm{y}+2\mathrm{b}\mathrm{z}=0$

Has now trivial solutions, then a, b, c are in

If $a, b, c$  are non - zero real numbers and if the system of equations

$\left(a-1\right)x=y+z$

$\left(b-1\right)y=x+z$

$\left(c-1\right)z=x+y$

has a non - trivial solution, then $ab+bc+ca$  equals :

An ordered pair $\left(\alpha , \beta \right)$ for which the system of linear equations
$\left(1+\alpha \right)x+\beta y+z=2$
$\alpha x+\left(1+\beta \right)y+z=3$
$\alpha x+\beta y+2z=2$

has a unique solution, is :

If the system of linear equations

$x+ky+3z=0$
$3x+ky-2z=0$
$2x+4y-3z=0$

has a non-zero solution $\left(x, y, z\right),$ then $\frac{xz}{y^2}$ is equal to:

The number of real values of $\lambda$ for which the system of linear equations, $2x+4y-\lambda z=0$, $4x+\lambda y+2z=0$ and $\lambda x+2y+2z=0$, has infinitely many solutions, is:

Let $\lambda$ be a real number for which the system of linear equations

$x+y+z=6,$

$4x+\lambda y-\lambda z=\lambda -2$ and

$3x+2y-4z=-5$

has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation: