Explanation

(b)
Applying $C_{1} \rightarrow C_{1} + C_{2} + C_{3}$ and taking common ($a + b + c)\$from $C_{1}$, we get
$(a + b + c)\ \left| \begin{array}{r}
1\ \ b\ \ c \\
1\ \ c\ \ a \\
1\ \ a\ \ b
\end{array} \right|$
Applying $R_{2} \rightarrow R_{2} - R_{1}$and $R_{3} \rightarrow R_{3} - R_{1}$, we get
$(a + b + c)\left| \begin{matrix}
1 & b & c \\
0 & c - b & a - c \\
0 & a - b & b - c
\end{matrix} \right|$
$= (a + b + c)\{ - (c - b)^{2} - (a - b)(a - c)\}$
$= - (a + b + c)\{ a^{2} + b^{2} + c^{2} - ab - bc - ca\}$
=$- \frac{1}{2}(a + b + c)\{ 2a^{2} + 2b^{2} + 2c^{2} - 2ab - 2bc - 2ac\}$
$= - \frac{1}{2}(a + b + c)\left\{ a - b \right)^{2} + (b - c)^{2} + (c - a)^{2}\}$
Which is always negative.