By using properties of determinants,show that:
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$

Let $\Delta = \left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3},$ we get:
$\Delta = \left|\begin{array}{ccc}a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
Taking $(a+b+c)$ common from $R_{1}$:
$\Delta = (a+b+c) \left|\begin{array}{ccc}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|$
Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$:
$\Delta = (a+b+c) \left|\begin{array}{ccc}1 & 0 & 0 \\ 2 b & -(a+b+c) & 0 \\ 2 c & 0 & -(a+b+c)\end{array}\right|$
Taking $(a+b+c)$ common from $C_{2}$ and $C_{3}$:
$\Delta = (a+b+c)(a+b+c)(a+b+c) \left|\begin{array}{ccc}1 & 0 & 0 \\ 2 b & -1 & 0 \\ 2 c & 0 & -1\end{array}\right|$
$\Delta = (a+b+c)^{3} [1((-1)(-1) - 0)] = (a+b+c)^{3}$
Hence,the result is proved.