What will be the value of x + y + z if cos-1 x + cos-1 y + cos-1 z = 3π?

The correct option is -3.

Let's understand the step-by-step mathematical reasoning to arrive at this answer.

We are given the equation:
$${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z=3\pi$$

Recall the range of the principal value branch of the inverse cosine function, ${\cos }^{-1}\theta$. For any real number $\theta$ in the domain $[-1,1]$, the value of ${\cos }^{-1}\theta$ is bounded as follows:
$$0\le {\cos }^{-1}\theta \le \pi$$

Since each term in our equation is an inverse cosine function, we have the individual inequalities:
$$0\le {\cos }^{-1}x\le \pi$$
$$0\le {\cos }^{-1}y\le \pi$$
$$0\le {\cos }^{-1}z\le \pi$$

Adding these three inequalities together gives the maximum possible value for their sum:
$${\cos }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}z\le \pi +\pi +\pi =3\pi$$

Because the sum is given to be exactly $3\pi$ (its absolute maximum possible value), each individual term in the sum must simultaneously be equal to its maximum value, $\pi$. Therefore, we have:
$${\cos }^{-1}x=\pi$$
$${\cos }^{-1}y=\pi$$
$${\cos }^{-1}z=\pi$$

Taking the cosine of both sides of each equation, we find the values of $x$, $y$, and $z$:
$$x=\cos \left(\pi \right)=-1$$
$$y=\cos \left(\pi \right)=-1$$
$$z=\cos \left(\pi \right)=-1$$

Now, we can substitute these values back into the expression $x+y+z$:
$$x+y+z=(-1)+(-1)+(-1)=-3$$

Thus, the value of $x+y+z$ is indeed $-3$.