If \(\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=(3 \pi / 2)\) then \(x^{10}+y^{10}+z^{10}+\left[3 /\left(x^{10}+y^{10}+z^{10}\right)\right]=\) (a) 0 (b) 2 (c) 4 (d) 3

We are given that: \(\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}\) Let \(A = \sin^{-1} x\), \(B = \sin^{-1} y\), and \(C = \sin^{-1} z\). Then we have: \(A + B + C = \frac{3\pi}{2}\) Since \(A, B, C\) are angles, we also know that \(\pi \leq A + B + C \leq 3\pi\).

Now, we apply the sine summation formula. We have: \(x = \sin A\) \(y = \sin B\) \(z = \sin C\)

Since we have the following relation: \(A + B + C = \frac{3\pi}{2}\) Using the sine addition formula: \(\cos (A + B) = -\cos(A) \cos(B) - \sin(A) \sin(B) = -\sin C\)

We have x, y, and z in terms of A, B, and C as follows: \(x = \sin A\) \(y = \sin B\) \(z = -\cos (A + B) = 1 - \sin A \sin B\)

Since the domain of sine inverse is between 0 and 1, we have: \(0 \leq x = \sin A \leq 1\) \(0 \leq y = \sin B \leq 1\) \(0 \leq z = 1 - \sin A \sin B \leq 1\)

Since x, y, z are in the range 0 to 1, x^10, y^10, z^10 must also be positive.

Now, let's find the value of the given expression: \(x^{10} + y^{10} + z^{10} + \frac{3}{x^{10} + y^{10} + z^{10}}\) = \( (\sin A)^{10} + (\sin B)^{10} + (1 - \sin A \sin B)^{10} + \frac{3}{(\sin A)^{10} + (\sin B)^{10} + (1 - \sin A \sin B)^{10}}\) Notice that the expression is in the form of \(a^2 + b^2 + c^2 + \frac{3}{a^2+b^2+c^2}\). Since a, b, and c are positive, their squares must also be positive. Therefore, the expression will have a minimum value greater than 0. On the other hand, it's clear that the expression can approach infinity. Therefore, \(x^{10} + y^{10} + z^{10} + \frac{3}{x^{10} + y^{10} + z^{10}}\) can only have the values \(2\) or \(3\). By substituting the corner values of the intervals for x, y, and z, we easily find that the only possible value is \(2\). Hence, the answer is (b) 2.