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(B) Let the given determinant be $D$. Multiply $R_1$ by $x$,$R_2$ by $y$,and $R_3$ by $z$:$D = \frac{1}{xyz} \left| {\begin{array}{*{20}{c}}{{x^4} + x

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$3$

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(B) Let the given determinant be $D$. Multiply $R_1$ by $x$,$R_2$ by $y$,and $R_3$ by $z$:$D = \frac{1}{xyz} \left| {\begin{array}{*{20}{c}}{{x^4} + x}&{{x^3}y}&{{x^3}z}\{x{y^3}}&{{y^4} + y}&{{y^3}z}\{x{z^3}}&{y{z^3}}&{{z^4} + z}\end{array}} ight| = 11$Taking $x, y, z$ common from $C_1, C_2, C_3$ respectively:$D = \frac{xyz}{xyz} \left| {\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^3}}&{{x^3}}\{{y^3}}&{{y^3} + 1}&{{y^3}}\{{z^3}}&{{z^3}}&{{z^3} + 1}\end{array}} ight| = 11$Applying $R_1 ightarrow R_1 + R_2 + R_3$:$D = (x^3 + y^3 + z^3 + 1) \left| {\begin{array}{*{20}{c}}1&1&1\{{y^3}}&{{y^3} + 1}&{{y^3}}\{{z^3}}&{{z^3}}&{{z^3} + 1}\end{array}} ight| = 11$Expanding the determinant,we get $(x^3 + y^3 + z^3 + 1)(1) = 11$,which implies $x^3 + y^3 + z^3 = 10$.Since $x, y, z$ are positive integers,the possible solutions $(x, y, z)$ are permutations of $(2, 1, 1)$.These are $(2, 1, 1), (1, 2, 1), (1, 1, 2)$.Thus,there are $3$ such solutions.

The number of positive integral solutions of the equation $\left| {\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^2}y}&{{x^2}z}\\{x{y^2}}&{{y^3} + 1}&{{y^2}z}\\{x{z^2}}&{y{z^2}}&{{z^3} + 1}\end{array}} \right| = 11$ is

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