Explanation

(b)
Let each ratio be $k$ and let $A = xyz$,
Then $\log{x = k(a - b),\ \ \log{y = k(b - c)}}$
And $\log{z = k(c - a)}$
$\therefore\ \ \ \log A = \log{x + \log{y + \log z}}$
$= k(a - b) + k(b - c) + k(c - a)$
$= k\lbrack a - b + b - c + c - a\rbrack$
$= k\lbrack 0\rbrack$
$\therefore\ \ \ \log{A = \log{(xyz) = 0\ \ \ \ \ \lbrack\because A = xyz\rbrack}}$
$\Rightarrow xyz = e^{0} = 1$