$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - {a^2}}&{ab}&{ac}\\{ba}&{ - {b^2}}&{bc}\\{ca}&{cb}&{ - {c^2}}\end{array}} \right| = 4{a^2}{b^2}{c^2}$

$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - {a^2}}&{ab}&{ac}\\{ba}&{ - {b^2}}&{bc}\\{ca}&{cb}&{ - {c^2}}\end{array}} \right|$

Taking a, b and c common from ${R_1},{R_2}$

and ${R_3}$respectively,

we get
$abc\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - a}&b&c\\a&{ - b}&c\\a&b&{ - c}\end{array}} \right|$

Applying ${R_1} \to {R_1} + {R_3},$

we get
$abc\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{2b}&0\\a&{ - b}&c\\a&b&{ - c}\end{array}} \right|$

Expanding along ${R_1}$,

we get
$(abc)( - 2b)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}a&c\\a&{ - c}\end{array}} \right|$

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

Hence, proved.