Explanation

(d)
We can write $\mathrm{\Delta} = \mathrm{\Delta}_{1} + y_{1}\mathrm{\Delta}_{2},$ where
$\mathrm{\Delta}_{1} = \left| \begin{matrix}
1 & 1 + x_{1}y_{2} & 1 + x_{1}y_{3} \\
1 & 1 + x_{2}y_{2} & 1 + x_{2}y_{3} \\
1 & 1 + x_{3}y_{2} & 1 + x_{3}y_{3}
\end{matrix} \right|$
and $\mathrm{\Delta}_{2} = \left| \begin{matrix}
x_{1} & 1 + x_{1}y_{2} & 1 + x_{1}y_{3} \\
x_{2} & 1 + x_{2}y_{2} & 1 + x_{2}y_{3} \\
x_{3} & 1 + x_{3}y_{2} & 1 + x_{3}y_{3}
\end{matrix} \right|$
In $\mathrm{\Delta}_{1}$, use $C_{2} \rightarrow C_{2} - C_{1}\ and\ C_{3} \rightarrow C_{3} - C_{1}$ so that,
$\mathrm{\Delta}_{1} = \left| \begin{matrix}
1 & x_{1}y_{2} & x_{1}y_{3} \\
1 & x_{2}y_{2} & x_{2}y_{3} \\
1 & x_{3}y_{2} & x_{3}y_{3}
\end{matrix} \right|$=0 $(\because C_{2}and\ C_{3}$ are proportional)
In $\mathrm{\Delta}_{2}$, $C_{2} \rightarrow C_{2} - {y_{2}C}_{1}$ and $C_{3} \rightarrow C_{3} - {y_{3}C}_{1}$ to get
$\mathrm{\Delta}_{2} = \left| \begin{matrix}
x_{1} & 1 & 1 \\
x_{2} & 1 & 1 \\
x_{3} & 1 & 1
\end{matrix} \right| = 0\ \ \ (\because C_{2}\ and\ C_{3}$ are identical)
$\therefore\ \mathrm{\Delta} = 0$