If $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\frac{3π}{2}$ and f(1) = 2, f(p + q) = f(p). f(q), ∀ p, q ∈ R, then the value of $x^{f(1)}+y^{f(2)}+z^{f(3)}-\frac{(x+y+z)}{x^{f(1)}+y^{f(2)}+z^{f(3)}}$ is:

2

$\sin^{-1}x=\sin^{-1}y=\sin^{-1}z=\frac{π}{2}$ or x = y = z = 1

f(1) = 2, f(2) = f(1 + 1) = f(1) . f(1) = 2 × 2 = 4, f(3) = f(2 + 1) = f(2) . f(1) = 4 × 2 = 8

$x^{f(1)}+y^{f(2)}+z^{f(3)}-\frac{(x+y+z)}{x^{f(1)}+y^{f(2)}+z^{f(3)}}=1^2+1^4+1^8-\frac{(1+1+1)}{1^2+1^4+1^8}=3-\frac{3}{3}=2$