​$$\text { The value of the determinant }\left|\begin{array}{lll}b^2-a b & b-c & b c-a c \\a b-a^2 & a-b & b^2-a b \\b c-a c & c-a & a b-a^2\end{array}\right|=\text { ? }$$​​

Correct option is C

​​$$\text{Given determinant:}\\[10pt]\Delta =\begin{vmatrix}b^2 - ab & b - c & bc - ac \\ab - a^2 & a - b & b^2 - ab \\bc - ac & c - a & ab - a^2\end{vmatrix}\\[12pt]\text{Factor } (b - a) \text{ from rows:}=\begin{vmatrix}b(b - a) & b - c & c(b - a) \\a(b - a) & a - b & b(b - a) \\c(b - a) & c - a & a(b - a)\end{vmatrix}\\[12pt]= (b - a)^2 \cdot\begin{vmatrix}b & b - c & c \\a & a - b & b \\c & c - a & a\end{vmatrix}\\[12pt]\text{Apply column operation: } C_1 \rightarrow C_1 - C_2\\[10pt]=(b - a)^2 \cdot\begin{vmatrix}b - (b - c) & b - c & c \\a - (a - b) & a - b & b \\c - (c - a) & c - a & a\end{vmatrix}\\[10pt]=(b - a)^2 \cdot\begin{vmatrix}c & b - c & c \\b & a - b & b \\a & c - a & a\end{vmatrix}\\[12pt]\text{Now, observe: } C_1 = C_3\\[10pt]\Rightarrow \text{Two identical columns} \Rightarrow \text{Determinant is zero}\\[12pt]\boxed{\Delta = 0}$$​​