The value of $\lambda$ for which the system of equations $2x - y - z = 12,$ $x - 2y + z = -4,$ and $x + y + \lambda z = 4$ has no solution is

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

If the system of equations $x+y+z=5$,$x+2y+2z=6$,and $x+3y+\lambda z=\mu$ (where $\lambda, \mu \in R$) is solvable by the Matrix Inversion Method,then:

Let $\alpha, \beta (\alpha \neq \beta)$ be the values of $m$ for which the equations $x+y+z=1$,$x+2y+4z=m$,and $x+4y+10z=m^2$ have infinitely many solutions. Then the value of $\sum_{n=1}^{10}(n^\alpha+n^\beta)$ is equal to:

The values of $a$ and $b$,for which the system of equations $2x + 3y + 6z = 8$,$x + 2y + az = 5$,and $3x + 5y + 9z = b$ has no solution,are:

If $p$ and $q$ are two distinct real values of $\lambda$ for which the system of equations $\begin{aligned} (\lambda-1) x+(3 \lambda+1) y+2 \lambda z &=0 \\ (\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z &=0 \\ 2 x+(3 \lambda+1) y+3(\lambda-1) z &=0 \end{aligned}$ has a non-zero solution,then $p^2+q^2-p q=$