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

Prove that $\left|\begin{array}{ccc}a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$

If $\left| {\begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} \\ {b - c}&{c - a}&{a - b} \\ {c - a + 1}&{a - b}&{b - c} \end{array}} \right| = 0$,where $a, b, c \in R - \{0\}$,then:

$A$ value of $\theta$ lying between $0$ and $\frac{\pi}{2}$ and satisfying $\left|\begin{array}{ccc} 1+\sin^2 \theta & \cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & 1+\cos^2 \theta & 4\sin 4\theta \\ \sin^2 \theta & \cos^2 \theta & 1+4\sin 4\theta \end{array}\right| = 0$ is:

Using properties of determinants,prove that:
$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.

If $\Delta = \begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix}$,then $\begin{vmatrix} ka & kb & kc \\ kx & ky & kz \\ kp & kq & kr \end{vmatrix}$ =