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

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(B) Given that $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$. We know that the range of $\cos ^{-1} 	heta$ is $[0, \pi]$. Since the sum of three val

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$x+y+z+3=0$

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(B) Given that $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$. We know that the range of $\cos ^{-1} 	heta$ is $[0, \pi]$. Since the sum of three values, each at most $\pi$, is $3 \pi$, each individual term must be equal to $\pi$. 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 $z = \cos \pi = -1$. Substituting these values into the expression $x+y+z+3$, we get $(-1) + (-1) + (-1) + 3 = -3 + 3 = 0$. Thus, $x+y+z+3=0$.

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$, then

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