If $a \ne 6, b, c$ satisfy $\left| \begin{array}{ccc} a & 2b & 2c \\ 3 & b & c \\ 4 & a & b \end{array} \right| = 0$,then $abc = $

The sum of all the roots of the equation $\left|\begin{array}{ccc}x & -3 & 2 \\ -1 & -2 & x-1 \\ 1 & x-2 & 3\end{array}\right|=0$ is

An equilateral triangle has each of its sides of length $6 \text{ cm}$. If $(x_1, y_1), (x_2, y_2), \text{ and } (x_3, y_3)$ are its vertices,then the value of the determinant $\left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right|^2$ is equal to:

If $\left|\begin{array}{ccc}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \neq 0$,then the area of the triangle whose vertices are $\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right), \left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right), \left(\frac{x_{3}}{c}, \frac{y_{3}}{c}\right)$ is:

Prove that $\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|=4abc$

If the system of equations $(\alpha + 1)^3 x + (\alpha + 2)^3 y - (\alpha + 3)^3 = 0$,$(\alpha + 1)x + (\alpha + 2)y - (\alpha + 3) = 0$,and $x + y - 1 = 0$ is consistent,what is the value of $\alpha$?