Explanation

(b)
Let $A\ and\ R$ be the first term and common ratio of the GP, then
$a = AR^{p - 1},b = AR^{q - 1}\ and = AR^{r - 1}\$ …(i)
Again, if $x$ and $d$ be the first term and common difference of an AP corresponding to the given HP, then
$\frac{1}{a} = x + (p - 1)d,\ \ \frac{1}{b} = x + (q - 1)d,\ \ \frac{1}{c} = x + (r - 1)d$ …(ii)
From Eq.(i), $\frac{a}{b} = R^{p - q}$
$\Rightarrow \left( \frac{a}{b} \right)^{1/c} = \left( R^{p - q} \right)^{1/c} = R^{k},$
Where $k = \frac{p - q}{c} = (p - q)\left\{ x + (r - 1)d \right\}$ [from Eq. (ii)]
$= (p - q)x + (p - q)(r - 1)d$
$= \ (p - q)x - (p - q)d + (rp - rq)d$ …(iii)
Similarly, $\left( \frac{b}{c} \right)^{1/a} = \left( R^{q - r} \right)^{1/a} = R^{n},$
Where $n = \frac{(q - r)}{a} = (q - r) \times \{ x + (p - 1)d\}$ [from Eq.(ii)]
$\Rightarrow n = (q - r)x - (q - r)d + (pq - pr)d$ …(iv)
And $\left( \frac{c}{a} \right)^{1/b} = \left( R^{r - p} \right)^{1/b} = R^{m}$
Where $m = \frac{r - p}{b} = (r - p)\left\{ x + (q - 1)d \right\}\$ [from Eq.(ii)]
$= (r - p)x(r - p)d + (rq - qp)d$ …(v)
$\therefore\left( \frac{a}{b} \right)^{1/c}\left( \frac{b}{c} \right)^{1/a}\ \left( \frac{c}{a} \right)^{1/b}$= $R^{k}R^{m}R^{n} = R^{m + n + k} = R^{0} = 1$
[Since, $k + m + n = 0,$ adding Eqs. (iii), (iv) and (v)]
Taking log on both sides, we get
$\frac{1}{c}(\log{a - \log{b) + \frac{1}{a}(\log{b - \log{c)}}}} + \frac{1}{b}(\log{c - \log{a) = 1\log{(1)}}}$
$\Rightarrow \left( \frac{1}{c} - \frac{1}{b} \right)\log{a + \left( \frac{1}{a} - \frac{1}{c} \right)\log{b + \left( \frac{1}{b} - \frac{1}{a} \right)\log{c = 0}}}$
$\Rightarrow \left( \frac{b - c}{bc} \right)\log{a + \left( \frac{c - a}{ac} \right)\log{b + \left( \frac{a - b}{ab} \right)\log{c = 0}}}$
$\Rightarrow a(b - c)\log{a + b(c - a)\log{b + c(a - b)\log{c = 0}}}$