Let $A = \begin{bmatrix} 3-t & 1 & 0 \\ -1 & 3-t & 1 \\ 0 & -1 & 0 \end{bmatrix}$ and $\det(A) = 5$, then find the value of $t$.

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $\mathbb{R},$ the determinant $\left| \begin{array}{ccc} y + 1 & \alpha & \beta \\ \alpha & y + \beta & 1 \\ \beta & 1 & y + \alpha \end{array} \right|$ is equal to

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

If $\omega$ is an imaginary root of unity,then the value of $\left| \begin{array}{ccc} a & b\omega^2 & a\omega \\ b\omega & c & b\omega^2 \\ c\omega^2 & a\omega & c \end{array} \right|$ is

Find the values of $k$ if the area of the triangle is $4$ square units and the vertices are $(-2, 0), (0, 4), (0, k)$.

If the system of homogeneous equations $\begin{aligned} & t x+(t+1) y+(t-1) z=0 \\ & (t+1) x+t y+(t+2) z=0 \\ & (t-1) x+(t+2) y+t z=0\end{aligned}$ in $x, y, z$ has a non-trivial solution, then $t$ is a root of the equation