If left| {begin{array}{*{20}{c}}{ - {a^2}}&{a | vedclass

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(C) Given determinant is $\Delta = \left| {\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\{ab}&{ - {b^2}}&{bc}\{ac}&{bc}&{ - {c^2}}\end{array}} ight|

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$4$

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(C) Given determinant is $\Delta = \left| {\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\{ab}&{ - {b^2}}&{bc}\{ac}&{bc}&{ - {c^2}}\end{array}} ight|$.Taking $a, b, c$ common from $R_1, R_2, R_3$ respectively:$\Delta = (abc) \left| {\begin{array}{*{20}{c}}{ - a}&b&c\a&{ - b}&c\a&b&{ - c}\end{array}} ight|$.Taking $a, b, c$ common from $C_1, C_2, C_3$ respectively:$\Delta = (abc)(abc) \left| {\begin{array}{*{20}{c}}{ - 1}&1&1\1&{ - 1}&1\1&1&{ - 1}\end{array}} ight| = {a^2}{b^2}{c^2} \left| {\begin{array}{*{20}{c}}{ - 1}&1&1\1&{ - 1}&1\1&1&{ - 1}\end{array}} ight|$.Now,expanding the determinant:$\left| {\begin{array}{*{20}{c}}{ - 1}&1&1\1&{ - 1}&1\1&1&{ - 1}\end{array}} ight| = -1(1 - 1) - 1(-1 - 1) + 1(1 - (-1)) = 0 + 2 + 2 = 4$.Therefore,$\Delta = 4{a^2}{b^2}{c^2}$.Comparing with $K{a^2}{b^2}{c^2}$,we get $K = 4$.

If $\left| {\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\\{ab}&{ - {b^2}}&{bc}\\{ac}&{bc}&{ - {c^2}}\end{array}} \right| = K{a^2}{b^2}{c^2},$ then $K = $

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