The system linear equations
$x-y+z=1$
$x+y-z=3$
$x-4y+4z=\alpha$ has

Q. The ordered triad $(x,y,z)$ which satisfies the system of linear equations is:
$x+y+z=6$
$x-y+z=2$
$3x+2y-4z=-5$

Q. The system of equations $\begin{array}{c}\alpha x+y+z=\alpha -1\\ x+\alpha y+z=\alpha -1\\ x+y+\alpha z=\alpha -1\end{array}$ has no solution, is $\alpha$ is

Q. Let $\lambda$ be a real number for which the system of linear equations
$x+y+z=6$
$4x+\lambda y-\lambda z=\lambda -2$
$3x+2y-4z=-5$
has infinitely many solutions. Then $\lambda$ is a root of the quadratic equation :

Q. The system of equations $3x-y+4z=2;x+2y-z=3;4x-y+\lambda z=-1$ has no solution. The value of $\lambda$ is

Q. If the system of linear equations
$\begin{array}{c}x-2y+kz=1\\ 2x+y+z=2\\ 3x-y-kz=3\end{array}$
has a solution $(x,y,z),z\ne 0$, then $(x,y)$ lies on the straight line whose equation is :