If cos ^{-1} x+cos ^{-1} y+cos ^{-1} z=3 pi,t | vedclass

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(D) We know that the range of $\cos ^{-1} 	heta$ is $[0, \pi]$.Given that $\cos ^{-1} x + \cos ^{-1} y + \cos ^{-1} z = 3\pi$.Since each term can be

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$5$

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(D) We know that the range of $\cos ^{-1} 	heta$ is $[0, \pi]$.Given that $\cos ^{-1} x + \cos ^{-1} y + \cos ^{-1} z = 3\pi$.Since each term can be at most $\pi$,the only way their sum can be $3\pi$ is if each term is exactly $\pi$.Therefore,$\cos ^{-1} x = \pi$,$\cos ^{-1} y = \pi$,and $\cos ^{-1} z = \pi$.This implies $x = \cos(\pi) = -1$,$y = \cos(\pi) = -1$,and $z = \cos(\pi) = -1$.Now,substitute these values into the expression $x^2 + y^2 + z^2 - 2xyz$:$(-1)^2 + (-1)^2 + (-1)^2 - 2(-1)(-1)(-1)$$= 1 + 1 + 1 - 2(-1)$$= 3 + 2$$= 5$.

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$,then the value of $x^2+y^2+z^2-2xyz$ is

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