If the system of equations $(\lambda-1) x+(\lambda-4) y+\lambda z=5$,$\lambda x+(\lambda-1) y+(\lambda-4) z=7$,and $(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9$ has infinitely many solutions,then $\lambda^2+\lambda$ is equal to

The system of equations $x+2y+3z=6$,$x+3y+5z=9$,and $2x+5y+az=12$ has no solution when $a=$

The solution of the equation $\begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$ is $(x, y, z) = $

Let $a$ be the sum of all coefficients in the expansion of $(1-2x+2x^2)^{2023}(3-4x^2+2x^3)^{2024}$ and $b = \lim_{x \rightarrow 0} \left( \frac{\int_0^x \frac{\ln(1+t)}{t^{2024}+1} dt}{x^2} \right)$. If the equations $cx^2+dx+e=0$ and $2bx^2+ax+4=0$ have a common root,where $c, d, e \in \mathbb{R}$,then $d:c:e$ equals

Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu$,and a system of linear equations
$x+y+z=5$
$x+2y+3z=\mu$
$x+3y+\lambda z=1$
is constructed. If $p$ is the probability that the system has a unique solution and $q$ is the probability that the system has no solution,then:

If the matrix equation is given,then $x=$ . . . . . . and $y=$ . . . . . . .