The number of solutions of the equations $x + 4y - z = 0,$ $3x - 4y - z = 0,$ and $x - 3y + z = 0$ is

The system of equations $(\sin\theta) x + 2z = 0$,$(\cos\theta) x + (\sin\theta) y = 0$,and $(\cos\theta) y + 2z = a$ has

If the system of linear equations $2x + 3y - z = -2$; $x + y + z = 4$; $x - y + |\lambda|z = 4\lambda - 4$ (where $\lambda \in R$) has no solution,then:

The number of solutions of the system of equations $2x + y - z = 7$,$x - 3y + 2z = 1$,and $x + 4y - 3z = 5$ is:

If the system of linear equations $x + 2ay + az = 0$,$x + 3by + bz = 0$,and $x + 4cy + cz = 0$ has a non-zero solution,then $a, b, c$:

If $x^a y^b=e^m, x^c y^d=e^n, \Delta_1=\left|\begin{array}{ll}m & b \\ n & d\end{array}\right|, \Delta_2=\left|\begin{array}{ll}a & m \\ c & n\end{array}\right|, \Delta_3=\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$,then the values of $x$ and $y$ are respectively ($e$ is the base of natural logarithm).