left| {begin{array}{*{20}{c}}{a - 1}&a&{bc}{b | vedclass

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(D) Let $\Delta = \left| {\begin{array}{*{20}{c}}{a - 1}&a&{bc}\{b - 1}&b&{ca}\{c - 1}&c&{ab}\end{array}} ight|$.Using the property of determinant

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(D) Let $\Delta = \left| {\begin{array}{*{20}{c}}{a - 1}&a&{bc}\{b - 1}&b&{ca}\{c - 1}&c&{ab}\end{array}} ight|$.Using the property of determinants,we can split the determinant as:$\Delta = \left| {\begin{array}{*{20}{c}}a&a&{bc}\b&b&{ca}\c&c&{ab}\end{array}} ight| - \left| {\begin{array}{*{20}{c}}1&a&{bc}\1&b&{ca}\1&c&{ab}\end{array}} ight|$.The first determinant is $0$ because the first two columns are identical.So,$\Delta = 0 - \left| {\begin{array}{*{20}{c}}1&a&{bc}\1&b&{ca}\1&c&{ab}\end{array}} ight| = - \left| {\begin{array}{*{20}{c}}1&a&{bc}\1&b&{ca}\1&c&{ab}\end{array}} ight|$.Multiply $R_1$ by $a$,$R_2$ by $b$,and $R_3$ by $c$ and divide by $abc$:$\Delta = - \frac{1}{abc} \left| {\begin{array}{*{20}{c}}a&{a^2}&{abc}\b&{b^2}&{abc}\c&{c^2}&{abc}\end{array}} ight| = - \frac{abc}{abc} \left| {\begin{array}{*{20}{c}}a&{a^2}&1\b&{b^2}&1\c&{c^2}&1\end{array}} ight| = - \left| {\begin{array}{*{20}{c}}a&{a^2}&1\b&{b^2}&1\c&{c^2}&1\end{array}} ight|$.Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\Delta = - \left| {\begin{array}{*{20}{c}}a&{a^2}&1\b-a&{b^2-a^2}&0\c-a&{c^2-a^2}&0\end{array}} ight| = - (b-a)(c-a) \left| {\begin{array}{*{20}{c}}a&{a^2}&1\1&{b+a}&0\1&{c+a}&0\end{array}} ight|$.Expanding along the third column:$\Delta = - (b-a)(c-a) [1 \cdot ((c+a) - (b+a))] = - (b-a)(c-a)(c-b) = (a-b)(b-c)(c-a)$.Since this result is not among the options $A, B, C$,the correct answer is $D$.

$\left| {\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}} \right| = $

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