If ${\Delta _1} = \left| {\begin{array}{*{20}{c}}1&0\\a&b\end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{*{20}{c}}1&0\\c&d\end{array}} \right|$,then ${\Delta _2}{\Delta _1}$ is equal to

$\left| {\begin{array}{*{20}{c}}a&b&c\\b&c&a\\c&a&b\end{array}} \right| = $

If $A = \begin{bmatrix} \alpha & 2 \\ 2 & \alpha \end{bmatrix}$ and $|A^3| = 27$,then $\alpha = $

If $a$ and $b$ are any two real numbers, then $\left|\begin{array}{ccc} 2a-2b-4 & 4a & 4a \\ 4 & 2-b-a & 4 \\ 2b & 2b & b-a-2 \end{array}\right| = $

The equation obtained by eliminating $a, b, c$ from the equations $x = \frac{a}{b-c}$,$y = \frac{b}{c-a}$,and $z = \frac{c}{a-b}$ is

Three-digit numbers $x17$,$3y6$,and $12z$,where $x, y, z$ are integers from $0$ to $9$,are divisible by a fixed constant $k$. Then the determinant $\left| \begin{array}{ccc} x & 3 & 1 \\ 7 & 6 & z \\ 1 & y & 2 \end{array} \right| + 48$ must be divisible by: