Let $\alpha, \beta \in R$ be such that the system of linear equations $x+2y+z=5, 2x+y+\alpha z=5, 8x+4y+\beta z=18$ has no solution. Then $\frac{\beta}{\alpha}$ is equal to:

Let $x = \alpha, y = \beta, z = \gamma$ be the unique solution of the system of simultaneous linear equations $2x + 3y - 2z + 4 = 0$, $3x - 4y + 3z + 5 = 0$, and $kx - 2y + z + 3 = 0$. If $\alpha = -2$, then $k =$

The positive value of $a$ for which the system of linear homogeneous equations $x+ay+z=0$, $ax+2y-z=0$, and $2x+3y+z=0$ has non-trivial solutions is

If $\frac{5}{m}+\frac{2}{n}=9$ and $\frac{3}{m}+\frac{4}{n}=11$ and $mn \neq 0$,then the values of $m$ and $n$ are . . . . . . respectively.

If the system of simultaneous linear equations $x+y+z=\lambda$,$5x-y+\mu z=10$,and $2x+3y-z=6$ has a unique solution,then:

The following system of equations $3x - 2y + z = 0$,$\lambda x - 14y + 15z = 0$,$x + 2y - 3z = 0$ has a solution other than $x = y = z = 0$ for $\lambda$ equal to