If sin−1⁡x+sin−1⁡y+sin−1⁡z=3π2 and f(1) = 1, f(p + q) = f(p). f(q)∀p, q ∈ R then, xf(1)+y(2)+zf(3)−x+y+zxf(1)+yf(2)+zf(3)=

Since, −π2≤sin−1⁡x≤π2∴sin−1⁡x+sin−1⁡y+sin−1⁡z=3π2⇒sin−1⁡x=sin−1⁡y=sin−1⁡z=π2⇒x=y=z=1Also, f(p + q) = f(p). f(q) ∀ p, q ∈ R      …. (1)Given, f(1) = 1From (1),f(1 + 1) = f(1). f(1)⇒ f(2) =12 = 1               ………. (2)From (2), f(2 + 1) = f(2) . f(1)⇒ f(3) = 12 .1 = 13 = 1Now, given expression =3−33=2