The system of linear equations $x + 2y + z = -3$,$3x + 3y - 2z = -1$,and $2x + 7y + 7z = -4$ has:

The system of equations $\begin{cases} \lambda x+y+3 z=0 \\ 2 x+\mu y-z=0 \\ 5 x+7 y+z=0 \end{cases}$ has infinitely many solutions in $\mathbb{R}$. Then,

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

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).

For a matrix $A=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,if $U_{1}, U_{2}$,and $U_{3}$ are $3 \times 1$ column matrices satisfying $A U_{1}=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$,$A U_{2}=\begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix}$,$A U_{3}=\begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$,and $U$ is a $3 \times 3$ matrix whose columns are $U_{1}, U_{2}$,and $U_{3}$,then the sum of the elements of $U^{-1}$ is:

If the system of equations
$ 11 x+y+\lambda z=-5 $
$ 2 x+3 y+5 z=3 $
$ 8 x-19 y-39 z=\mu $
has infinitely many solutions,then $ \lambda^4-\mu $ is equal to :