Explanation

(a)
In $\mathrm{\Delta}ABC$, given $\left| \begin{matrix}
1 & a & b \\
1 & c & a \\
1 & b & c
\end{matrix} \right| = 0$
$\Rightarrow 1\left( c^{2} - ab \right) - a(c - a) + b(b - c) = 0$
$\Rightarrow a^{2} + b^{2} + c^{2} - ab - bc - ca = 0$
$\Rightarrow \ {2a}^{2} + 2b^{2} + {2c}^{2} - 2ab - 2bc - 2ca = 0$
$\Rightarrow \left( a^{2} + b^{2} - 2ab \right) + \left( b^{2} + c^{2} - 2bc \right) + \left( c^{2} + a^{2} - 2ca \right) = 0$
$\Rightarrow (a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 0$
Here, sum of squares of three numbers can be zero, if and only, if $a = b = c$.
$\Rightarrow \ \mathrm{\Delta}\ ABC$ is an equilateral triangle.
$\Rightarrow \ \angle A = \angle B = \angle C = 60{^\circ}$
$\therefore\sin^{2}{A + \sin^{2}{B + \sin^{2}{C = \sin^{2}{60{^\circ} + \sin^{2}{60{^\circ} + \sin^{2}{60{^\circ}}}}}}}$
$= \left( \frac{3}{4} + \frac{3}{4} + \frac{3}{4} \right) = \frac{9}{4}$