Using properties of determinants, prove that:

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

Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$, 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 $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}$, we have:

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

Expanding along $\mathrm{C}_{1}$, we have:

$\begin{aligned} \Delta &=(a+b+c)[(2 b+a)(2 c+a)-(a-b)(a-c)] \\ &=(a+b+c)\left[4 b c+2 a b+2 a c+a^{2}-a^{2}+a c+b a-b c\right] \\ &=(a+b+c)(3 a b+3 b c+3 a c) \\ &=3(a+b+c)(a b+b c+c a) \end{aligned}$

Hence, the given result is proved.