Explanation

(c)
$\because - \frac{\pi}{2} \leq \sin^{- 1}{x \leq \frac{\pi}{2}, - \frac{\pi}{2} \leq \sin^{- 1}{y \leq \frac{\pi}{2}}}$
And $- \frac{\pi}{2} \leq \sin^{- 1}{\mathcal{z \leq}\frac{\pi}{2}}$
Given that, $\sin^{- 1}{x + \sin^{- 1}{y + \sin^{- 1}{\mathcal{z =}\frac{3\pi}{2}}}}$
Which is possible only when
$\sin^{- 1}{x = \sin^{- 1}{y = \sin^{- 1}{\mathcal{z =}\frac{\pi}{2}}}}$
Or $x = y = \mathcal{z} = 1$
Put $p = q = 1$
Then $f(2) = f(1)f(1) = 2 \bullet 2 = 4$
And put $p = 1,q = 2$
Then, $f(3) = f(1)f(2) = {2 \bullet 2}^{2} = 8$
$\therefore x^{f(1)} + y^{f(2)} + \mathcal{z}^{f(3)} - \frac{x + y + \mathcal{z}}{x^{f(1)} + y^{f(2)} + \mathcal{z}^{f(3)}}$
$= 1 + 1 + 1 - \frac{3}{1 + 1 + 1}$
$= 3 - 1 = 2$