The determinant equals

Option (D)

Given $\left|\begin{array}{lll}b^{2}-a b & b-c & b c-a c \\ a b-a^{2} & a-b & b^{2}-a b \\ b c-a c & c-a & a b-a^{2}\end{array}\right|$

$=\left|\begin{array}{lll}b(b-a) & b-c & c(b-a) \\ a(b-a) & a-b & b(b-a) \\ c(b-a) & c-a & a(b-a)\end{array}\right|$

Now, [Taking $(b-a)$ common from $C_{1}$ and $C_{3}$ each]

$=(b-a)^{2}\left|\begin{array}{lll}b & b-c & c \\ a & a-b & b \\ c & c-a & a\end{array}\right|$

[Applying $C_{2} \rightarrow C_{2}+C_{3}$ ]

$=(b-a)^{2}\left|\begin{array}{lll}b & b & c \\ a & a & b \\ c & c & a\end{array}\right|$

$=0$   [as $C_{1}$ and $C_{2}$ are identical]