If x, y, z ∈ [-1,1] such that $(sin^{-1} x +sin^{-1} y+ sin^{-1} z) = -\frac{3\pi}{2},$ then $ x^2 + y^2 + z^2 = $

3

We know that the minimum value of $sin^{-1} x$ for $ x ∈ [-1,1] $ is $-\frac{\pi}{2}$.

$∴ sin^{-1} x ≥ -\frac{\pi}{2}, sin^{-1} y ≥-\frac{\pi}{2} \, \, and \,\, sin^{-1} z ≥ -\frac{\pi}{2}$ for all $ x, y, z ∈ [-1,1]$

$⇒ sin^{-1}x + sin^{-1} y + sin^{-1} z ≥ \left(-\frac{\pi}{2}\right)+\left(-\frac{\pi}{2}\right)+\left(-\frac{\pi}{2}\right)$

$⇒ sin^{-1}x + sin^{-1} y + sin^{-1} z ≥ -\frac{3\pi}{2}$

$∴  sin^{-1}x + sin^{-1} y + sin^{-1} z = -\frac{3\pi}{2}$

$⇒sin^{-1}x =-\frac{\pi}{2}, sin^{-1}y = -\frac{\pi}{2}, sin^{-1}z =-\frac{\pi}{2}$

$⇒ x = y = z = -1$

Hence, $x^2 + y^2 + z^2= (-1)^2 + (-1)^2 + (-1)^2= 3$