The system of linear equations $x+y+z=6, x+2y+3z=10$ and $x+2y+az=b$ has no solutions when

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

If the system of equations $\alpha x + y + z = 5$,$x + 2y + 3z = 4$,and $x + 3y + 5z = \beta$ has infinitely many solutions,then the ordered pair $(\alpha, \beta)$ is equal to:

The following system of equations $3x - 2y + z = 0$,$\lambda x - 14y + 15z = 0$,$x + 2y - 3z = 0$ has a solution other than $x = y = z = 0$ for $\lambda$ equal to

For what values of $a$ will the system of equations $x+y+z=1$, $2x+3y+2z=2$, and $ax+ay+2az=4$ have a unique solution?

If the system of linear equations $x + ky + 3z = 0$,$3x + ky - 2z = 0$,and $2x + 4y - 3z = 0$ has a non-zero solution $(x, y, z)$,then $\frac{xz}{y^2} = \dots$