The determinant $\left|\begin{array}{lll}b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-c a & c-a & a b-a^{2}\end{array}\right|$ equals
(a) $a b c(b-c)(c-a)(a-b)$
(b) $(b-c)(c-a)(a-b)$
(c) $(a+b+c)(b-c)(c-a)(a-b)$
(d) none of these
$\left|\begin{array}{lll}b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-c a & c-a & a b-a^{2}\end{array}\right|$
$=\left|\begin{array}{lll}b(b-a) & b-c & c(b-a) \\ a(b-a) & a-b & b(b-a) \\ c(b-a) & c-a & a(b-a)\end{array}\right|$
$=(b-a)^{2}\left|\begin{array}{lll}b & b-c & c \\ a & a-b & b \\ c & c-a & a\end{array}\right|$ [Taking $(b-a)$ common from $C_{1}$ and $C_{3}$ ]
$=(b-a)^{2}\left|\begin{array}{lll}0 & b-c & c \\ 0 & a-b & b \\ 0 & c-a & a\end{array}\right|$ [Applying $C_{1} \rightarrow C_{1}-C_{2}-C_{3}$ ]
$=0$
Hence, the correct option is $(\mathrm{d})$.
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