If $\left| \begin{array}{ccc} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & b+c & -2c \end{array} \right| = \alpha (a+b)(b+c)(c+a) \neq 0$,then $\alpha$ is equal to

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

Let $A$ be a $3 \times 3$ matrix with $\operatorname{det}(A) = 4$. Let $R_{i}$ denote the $i^{\text{th}}$ row of $A$. If a matrix $B$ is obtained by performing the operation $R_{2} \rightarrow 2R_{2} + 5R_{3}$ on $2A$,then $\operatorname{det}(B)$ is equal to:

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

By using properties of determinants,show that:
$\left|\begin{array}{ccc}0 & a & -b \\ -a & 0 & -c \\ b & c & 0\end{array}\right|=0$

$\left| \begin{array}{ccc} 13 & 16 & 19 \\ 14 & 17 & 20 \\ 15 & 18 & 21 \end{array} \right| = $