If $ax^4+bx^3+cx^2+50x+d = \begin{vmatrix} x^3-14x^2 & -x & 3x+\lambda \\ 4x+1 & 3x & x-4 \\ -3 & 4 & 0 \end{vmatrix}$,then find $\lambda$.

If points $(5, 5)$,$(10, k)$ and $(-5, 1)$ are collinear,then $k =$

$\begin{aligned} & \text { If }\left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2 & (n+4)^2\end{array}\right|=\Delta \text { and } \\ & \left|\begin{array}{ccc}1 & -4 & 7 \\ -2 & 3 & -5 \\ 3 & x & -3\end{array}\right|=2 \Delta+1, \text { then } x=\end{aligned}$

If $D = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{array} \right|$ for $x \neq 0, y \neq 0$,then $D$ is

If $x = cy + bz$,$y = az + cx$,$z = bx + ay$ (where $x, y, z$ are not all zero) have a solution other than $x = 0, y = 0, z = 0$,then $a, b$,and $c$ are connected by the relation:

If $\left[\begin{array}{rrr}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$ has no inverse, then the real value of $x$ is