Explanation

(b)
Given equation is
$\left( p^{2} + q^{2} \right)x^{2} - 2q(p + r)x + (q^{2} + r^{2}) = 0$
Since, roots are real and equal, then
$b^{2} - 4ac = 0$
$\Rightarrow 4q^{2}(p + r)^{2} - 4\left( p^{2} + q^{2} \right)\left( q^{2} + r^{2} \right) = 0$
$\Rightarrow q^{2}\left( p^{2} + r^{2} + 2pr \right) - \left( p^{2}q^{2} + p^{2}r^{2} + q^{4} + q^{2}r^{2} \right) = 0$
$\Rightarrow q^{2}p^{2} + q^{2}r^{2} + 2pq^{2}r - p^{2}q^{2} - p^{2}r^{2} - q^{4} - q^{2}r^{2} = 0$
$\Rightarrow 2pq^{2}r - p^{2}r^{2} - q^{4} = 0$
$\Rightarrow \left( q^{2} - pr \right)^{2} = 0$
$\Rightarrow q^{2} = pr$
$\therefore p,q\ and\ r$ will be in GP.