Let $A$ be a square matrix of order $3 \times 3$,then $|5A| = $ (in $|A|$)

By using properties of determinants,show that:
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$

$2\,\,\left| {\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ {a^2 - bc} & {b^2 - ac} & {c^2 - ab} \end{array}} \right| = $

If $x, y, z \in \mathbb{R}$,then the value of the determinant $\left|\begin{array}{lll}\left(5^{x}+5^{-x}\right)^{2} & \left(5^{x}-5^{-x}\right)^{2} & 1 \\ \left(6^{x}+6^{-x}\right)^{2} & \left(6^{x}-6^{-x}\right)^{2} & 1 \\ \left(7^{x}+7^{-x}\right)^{2} & \left(7^{x}-7^{-x}\right)^{2} & 1\end{array}\right|$ is:

$\left| {\begin{array}{ccc} a + b & b + c & c + a \\ b + c & c + a & a + b \\ c + a & a + b & b + c \end{array}} \right| = K \left| {\begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}} \right|$,then $K = $

Using properties of determinants,prove that:
$\left| \begin{array}{ccc} \sin \alpha & \cos \alpha & \cos (\alpha + \delta) \\ \sin \beta & \cos \beta & \cos (\beta + \delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma + \delta) \end{array} \right| = 0$