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$2\,\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2} - bc}&{{b^2} - ac}&{{c^2} - ab}\end{array}\,} \right| = $
$0$
$1$
$2$
$3abc$
Solution
(a) We have $2\,\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2} – bc}&{{b^2} – ac}&{{c^2} – ab}\end{array}\,} \right|$
= $2\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\end{array}\,} \right| – 2\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{bc}&{ac}&{ab}\end{array}\,} \right|$
= $2\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\end{array}\,} \right| – \frac{2}{{abc}}\left| {\,\begin{array}{*{20}{c}}a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\\{abc}&{abc}&{abc}\end{array}\,} \right|$
{ Applying ${C_1}(a),{C_2}(b),{C_3}(c)$}
$ = 2\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\end{array}\,} \right| – \frac{2}{{abc}}(abc)\,\left| {\,\begin{array}{*{20}{c}}a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\\1&1&1\end{array}\,} \right| = 0$.
