Explanation

(d)
Applying $R_{1} \rightarrow R_{1} + R_{2} + R_{3}\$and taking common ($a + b + c)\$from $R_{1},$we get
$= (a + b + c)\left| \begin{matrix}
1 & 1 & 1 \\
2b & b - c - a & 0 \\
2c & 0 & c - a - b
\end{matrix} \right|$
Applying $C_{2} \rightarrow C_{2} - C_{1}$ and $C_{3} \rightarrow C_{3} - C_{1}$,
$= (a + b + c)\left| \begin{matrix}
1 & 0 & 0 \\
2b & - b - c - a & - 2b \\
2c & 0 & - a - b - c
\end{matrix} \right|$
$= (c + b + c)\lbrack( - b - c - a)( - a - b - c)\rbrack$
$= (a + b + c)^{3}$