If the system of linear equations $x+y+3z=0$,$x+3y+k^{2}z=0$,and $3x+y+3z=0$ has a non-zero solution $(x, y, z)$ for some $k \in R$,then $x + (y/z)$ is equal to

If the system of equations $x+y+z=1$,$x+2y+4z=k$ and $x+4y+10z=k^2$ is consistent,then $k$ is equal to

The following system of linear equations $7x + 6y - 2z = 0$; $3x + 4y + 2z = 0$; $x - 2y - 6z = 0$ has:

Let $X = \begin{bmatrix} a \\ b \\ c \end{bmatrix}$,$A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{bmatrix}$,and $B = \begin{bmatrix} 3 \\ 1 \\ 4 \end{bmatrix}$. If $AX = B$,then the value of $2a - 3b + 4c$ is:

All the real values of $p, q$ so that the system of equations $\begin{cases} 2x + py + 6z = 8 \\ x + 2y + qz = 5 \\ x + y + 3z = 4 \end{cases}$ may have no solution are

Consider the system of equations: $ax + by + cz = 2$,$bx + cy + az = 2$,$cx + ay + bz = 2$,where $a, b, c$ are real numbers such that $a + b + c = 0$. Then,the system