If $A = \begin{bmatrix} 5 & 5\alpha & \alpha \\ 0 & \alpha & 5\alpha \\ 0 & 0 & 5 \end{bmatrix}$ and $\operatorname{det}(A^2) = 25$, then $|\alpha| = $

The determinant $\left| \begin{array}{ccc} a & b & a\alpha + b \\ b & c & b\alpha + c \\ a\alpha + b & b\alpha + c & 0 \end{array} \right| = 0$,if $a, b, c$ are in

If ${a^2} + {b^2} + {c^2} + ab + bc + ca \leq 0$ for all $a, b, c \in R$,then find the value of the determinant $\left| {\begin{array}{*{20}{c}} {{(a + b + c)}^2} & {{a^2} + {b^2}} & 1 \\ 1 & {{(b + c + 2)}^2} & {{b^2} + {c^2}} \\ {{c^2} + {a^2}} & 1 & {{(c + a + 2)}^2} \end{array}} \right|$.

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \alpha & \beta \\ 0 & \beta & \alpha \end{bmatrix}$ and $|2A|^3 = 2^{21}$,where $\alpha, \beta \in \mathbb{Z}$. Then a value of $\alpha$ is:

If $a, b, c$ are non-zero real numbers and the system of equations $(a - 1)x = y + z,$ $(b - 1)y = z + x,$ $(c - 1)z = x + y$ has a non-trivial solution,then $ab + bc + ca$ equals

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: