If $\omega$ is an imaginary cube root of unity and $\left|\begin{array}{ccc}x+\omega^2 & \omega & 1 \\ \omega & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2\end{array}\right|=0$, then one of the values of $x$ is

The area of a triangle is $5$. If two of its vertices are $(2, 1)$ and $(3, -2)$ and the third vertex lies on the line $y = x + 3$,then the third vertex is

If $\left|\begin{array}{cc}x^3+2 x^2+3 x-2 & x^2+2 x+4 \\ x^3-x^2-2 x-1 & 3 x^3-2 x^2+4 x-2\end{array}\right| = a x^6+b x^5+c x^4+d x^3+e x^2+f x+g$,then $a+b+c+d+e+f$ is equal to

$\left| \begin{matrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{matrix} \right| = $

Let $\alpha, \beta, \gamma$ be the real roots of the equation $x^{3} + ax^{2} + bx + c = 0$,where $a, b, c \in R$ and $a, b \neq 0$. If the system of equations in $u, v, w$ given by $\alpha u + \beta v + \gamma w = 0$,$\beta u + \gamma v + \alpha w = 0$,and $\gamma u + \alpha v + \beta w = 0$ has a non-trivial solution,then the value of $\frac{a^{2}}{b}$ is:

The matrix $\begin{bmatrix} 1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{bmatrix}$ is not invertible if $a$ has the value: