(d)
Given, $f(x) = \left| \begin{matrix}
1 + a & 1 + ax & 1 + ax^{2} \\
1 + b & 1 + bx & 1 + {bx}^{2} \\
1 + b & 1 + cx & 1 + {cx}^{2}
\end{matrix} \right|$
$\Rightarrow f(x) = \left| \begin{matrix}
1 + a & a(x - 1) & ax(x - 1) \\
1 + b & b(x - 1) & bx(x - 1) \\
1 + b & c(x - 1) & cx(x - 1)
\end{matrix} \right|$
$= (x - 1)x(x - 1)\left| \begin{array}{r}
1 + a\ \ a\ \ a \\
1 + b\ \ b\ \ b \\
1 + c\ \ c\ \ c
\end{array} \right|$=0
($\therefore two\ columns\ are\ same)$