Using properties of determinants,prove that:
$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

(A) Let $\Delta = \left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|$.
Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$,we get:
$\Delta = \left|\begin{array}{ccc}a+b+c & -a+b & -a+c \\ a+b+c & 3 b & -b+c \\ a+b+c & -c+b & 3 c\end{array}\right|$.
Taking $(a+b+c)$ common from $C_{1}$:
$\Delta = (a+b+c) \left|\begin{array}{ccc}1 & -a+b & -a+c \\ 1 & 3 b & -b+c \\ 1 & -c+b & 3 c\end{array}\right|$.
Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$:
$\Delta = (a+b+c) \left|\begin{array}{ccc}1 & -a+b & -a+c \\ 0 & 2 b+a & a-b \\ 0 & a-c & 2 c+a\end{array}\right|$.
Expanding along $C_{1}$:
$\Delta = (a+b+c) [ (2 b+a)(2 c+a) - (a-b)(a-c) ]$.
Simplifying the expression inside the bracket:
$= (a+b+c) [ (4bc + 2ab + 2ac + a^2) - (a^2 - ac - ab + bc) ]$
$= (a+b+c) [ 4bc + 2ab + 2ac + a^2 - a^2 + ac + ab - bc ]$
$= (a+b+c) [ 3ab + 3bc + 3ac ]$
$= 3(a+b+c)(ab + bc + ca)$.
Hence,the result is proved.