left| {begin{array}{*{20}{c}}{{b^2} + {c^2}}& | vedclass

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Question

(C) Let $\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

Vedclass · Accepted Answer

$4{a^2}{b^2}{c^2}$

Vedclass · Answer

(C) Let $\Delta = \left| {\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}} ight|$.Apply the operation ${R_1} 	o {R_1} - ({R_2} + {R_3})$:$\Delta = \left| {\begin{array}{*{20}{c}}{-2{c^2}}&{-2{c^2}}&{0}\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}} ight|$.Alternatively,taking common factors from columns or rows:$\Delta = 2{a^2}{b^2}{c^2} \left| {\begin{array}{*{20}{c}}{\frac{{{b^2} + {c^2}}}{{{a^2}}}}&1&1\1&{\frac{{{c^2} + {a^2}}}{{{b^2}}}}&1\1&1&{\frac{{{a^2} + {b^2}}}{{{c^2}}}}\end{array}} ight|$.By simplifying the determinant,we get:$\Delta = 4{a^2}{b^2}{c^2}$.Trick: Put $a=1, b=1, c=1$ is not ideal as options might overlap. Let $a=1, b=2, c=3$:$\Delta = \left| {\begin{array}{*{20}{c}}{13}&1&1\4&10&4\9&9&5\end{array}} ight| = 13(50-36) - 1(20-36) + 1(36-90) = 13(14) + 16 - 54 = 182 + 16 - 54 = 144$.Checking options: $4{a^2}{b^2}{c^2} = 4(1)^2(2)^2(3)^2 = 4 	imes 1 	imes 4 	imes 9 = 144$. Hence,option $C$ is correct.

$\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| = $

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