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:

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).

The system $x+2y+3z=4$,$4x+5y+3z=5$,$3x+4y+3z=\lambda$ is consistent and $3\lambda=n+100$,then $n=$

For how many different values of $a$ does the following system of linear equations have at least two distinct solutions?
$ax + y = 0$
$x + (a + 10)y = 0$

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 $k =$

If the system of equations $2x - y + z = 4$,$5x + \lambda y + 3z = 12$,and $100x - 47y + \mu z = 212$ has infinitely many solutions,then $\mu - 2\lambda$ is equal to