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$\left| {\,\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right| = $
$abc$
$4abc$
$4{a^2}{b^2}{c^2}$
${a^2}{b^2}{c^2}$
Solution
(c) $\Delta = \left| {\,\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right|$
=$ – 2\,\left| {\,\begin{array}{*{20}{c}}0&{{c^2}}&{{b^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right|$,by ${R_1}→{R_1}-(R_2 + R_3)$
= $ – 2\,\left| {\,\begin{array}{*{20}{c}}0&{{c^2}}&{{b^2}}\\{{b^2}}&{{a^2}}&0\\{{c^2}}&0&{{a^2}}\end{array}\,} \right|$, by $\begin{array}{l}{R_2} \to {R_2} – {R_1}\\{R_3} \to {R_3} – {R_1}\end{array}$
= $ – 2\{ – {c^2}({b^2}{a^2}) + {b^2}( – {c^2}{a^2})\} = 4{a^2}{b^2}{c^2}$.
Trick: Put $a=1,b=2,c=3$ so that the option give different values.
