= $\frac{{{a^2}{b^2}{c^2}}}{{abc}}\,\left| {\,\begin{array}{*{20}{c}}{bc}&1&{ab + ac}\\{ac}&1&{bc + ab}\\{ab}&1&{ac + bc}\end{array}\,} \right|\, = \,abc\,\left| {\,\begin{array}{*{20}{c}}{bc}&1&{\Sigma ab}\\{ac}&1&{\Sigma \,ab}\\{ab}&1&{\Sigma \,ab}\end{array}\,} \right|$

{by ${C_3} \to {C_3} + {C_1}$}

= $abc.\Sigma \,ab\,\left| {\,\begin{array}{*{20}{c}}{bc}&1&1\\{ca}&1&1\\{ab}&1&1\end{array}\,} \right| = 0$,          [Since ${C_2} \equiv {C_3}$].

Trick : Put $a = 1,\,b = 2,\,c = 3$ and check it.