Using properties of determinants, prove that:

$\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$

$\Delta=\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right|$

Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3} .$ We have:

$\Delta=\left|\begin{array}{ccc}a+b+c & -a+b & -a+c \\ a+b+c & 3 b & -b+c \\ a+b+c & -c+b & 3 c\end{array}\right|$

$=(a+b+c)\left|\begin{array}{ccc}1 & -a+b & -a+c \\ 1 & 3 b & -b+c \\ 1 & -c+b & 3 c\end{array}\right|$

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1},$ we have:

$\Delta=(a+b+c)\left|\begin{array}{ccc}1 & -a+b & -a+c \\ 0 & 2 b+a & a-b \\ 0 & a-c & 2 c+a\end{array}\right|$

Expanding along $C_{1},$ we have:

$\Delta=(a+b+c)[(2 b+a)(2 c+a)-(a-b)(a-c)]$

$=(a+b+c)\left[4 b c+2 a b+2 a c+a^{2}-a^{2}+a c+b a-b c\right]$

$=(a+b+c)(3 a b+3 b c+3 a c)$

$=3(a+b+c)(a b+b c+c a)$

Hence, the given result is proved.