Let the system of linear equations $x+y+kz=2$; $2x+3y-z=1$; $3x+4y+2z=k$ have infinitely many solutions. Then the system $(k+1)x+(2k-1)y=7$; $(2k+1)x+(k+5)y=10$ has:

The values of $p$ and $q$ so that the system of equations $2x + py + 6z = 8$,$x + 2y + qz = 5$ and $x + y + 3z = 4$ may have no solution are

Let $A = \begin{bmatrix} 1 & 3 \\ 4 & -3 \end{bmatrix}$. Let $S = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb{R}^2 \mid A \begin{bmatrix} x \\ y \end{bmatrix} = 3 \begin{bmatrix} x \\ y \end{bmatrix} \right\}$. What is the cardinality of $S$?

Let the system of linear equations $x+y+\alpha z=2$,$3x+y+z=4$,and $x+2z=1$ have a unique solution $(x^{*}, y^{*}, z^{*})$. If $(\alpha, x^{*}), (y^{*}, \alpha)$ and $(x^{*}, -y^{*})$ are collinear points,then the sum of absolute values of all possible values of $\alpha$ is

The system of equations $kx + 2y - z = 1$,$(k - 1)y - 2z = 2$,and $(k + 2)z = 3$ has a unique solution if $k$ is equal to:

If the system of linear equations : $x+y+2z=6$,$2x+3y+az=a+1$,$-x-3y+bz=2b$ where $a, b \in R$,has infinitely many solutions,then $7a+3b$ is equal to :