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

For non-zero,real $a, b$ and $c$,if $\left| \begin{array}{ccc} \frac{a^2+b^2}{c} & c & c \\ a & \frac{b^2+c^2}{a} & a \\ b & b & \frac{c^2+a^2}{b} \end{array} \right| = \alpha abc$,then the value of $\alpha$ is

The value of the determinant $\left| \begin{array}{ccc} 2 & 8 & 4 \\ -5 & 6 & -10 \\ 1 & 7 & 2 \end{array} \right|$ is

If $a - 2b + c = 1$,then the value of $\left| \begin{array}{ccc} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} \right|$ is

If $A$ and $B$ are two square matrices with $\det(A) = 5$ and $\det(B^T \cdot A^T) = -15$,then $\det(B)$ is equal to

Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$