If left| {begin{array}{*{20}{c}}{{{(b + c)}^2 | vedclass

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(B) Given determinant $\Delta = \left| {\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2}}&{{a^2}}\{{b^2}}&{{{(c + a)}^2}}&{{b^2}}\{{c^2}}&{{c^2}}&{{{(a

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(B) Given determinant $\Delta = \left| {\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2}}&{{a^2}}\{{b^2}}&{{{(c + a)}^2}}&{{b^2}}\{{c^2}}&{{c^2}}&{{{(a + b)}^2}}\end{array}} ight|$.Applying $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$:$\Delta = \left| {\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2} - {{(b + c)}^2}}&{{a^2} - {{(b + c)}^2}}\{{b^2}}&{{{(c + a)}^2} - {b^2}}&0\{{c^2}}&0&{{{(a + b)}^2} - {c^2}}\end{array}} ight|$$= \left| {\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{(a - b - c)(a + b + c)}&{(a - b - c)(a + b + c)}\{{b^2}}&{(c + a - b)(c + a + b)}&0\{{c^2}}&0&{(a + b - c)(a + b + c)}\end{array}} ight|$Taking out $(a + b + c)$ from $C_2$ and $C_3$:$\Delta = {(a + b + c)^2} \left| {\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{a - b - c}&{a - b - c}\{{b^2}}&{c + a - b}&0\{{c^2}}&0&{a + b - c}\end{array}} ight|$Applying $R_1 	o R_1 - R_2 - R_3$:$\Delta = {(a + b + c)^2} \left| {\begin{array}{*{20}{c}}{2bc}&{ - 2c}&{ - 2b}\{{b^2}}&{c + a - b}&0\{{c^2}}&0&{a + b - c}\end{array}} ight|$Expanding along $R_1$:$\Delta = {(a + b + c)^2} [2bc((c + a - b)(a + b - c)) + 2c(b^2(a + b - c)) - 2b(b^2(c + a - b)) ]$After simplification,we get $\Delta = 2abc{(a + b + c)^3}$.Comparing with $k\,abc{(a + b + c)^3}$,we get $k = 2$.

If $\left| {\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{{(c + a)}^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{{(a + b)}^2}}\end{array}} \right| = k\,abc{(a + b + c)^3}$,then the value of $k$ is

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