1. Given conditions

Since $x,y,z$ are in A.P., we have $2y=x+z.$

Also, ${\tan^{-1}}x, {\tan^{-1}}y, {\tan^{-1}}z$ are in A.P., so $2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z.$

Let $\alpha=\tan^{-1}x,\quad \beta=\tan^{-1}y,\quad \gamma=\tan^{-1}z.$ Then $2\beta=\alpha+\gamma.$ So $\beta$ is the arithmetic mean of $\alpha$ and $\gamma$.

2. Use tangent on the inverse-trigonometric A.P. condition

From $2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z,$ we take tangent on both sides: $\tan(2\tan^{-1}y)=\tan(\tan^{-1}x+\tan^{-1}z).$

Now, $\tan(2\tan^{-1}y)=\frac{2y}{1-y^2},$

and $\tan(\tan^{-1}x+\tan^{-1}z)=\frac{x+z}{1-xz}.$

Hence, $\frac{2y}{1-y^2}=\frac{x+z}{1-xz}.$

But from A.P. of $x,y,z$, $x+z=2y.$ So $\frac{2y}{1-y^2}=\frac{2y}{1-xz}.$

Thus, $2y\left(\frac{1}{1-y^2}-\frac{1}{1-xz}\right)=0.$

So either

• $y=0$, or
• $1-y^2=1-xz$, i.e. $y^2=xz$.

3. Combine with A.P. condition

We already know $x+z=2y.$ If also $xz=y^2$, then $x,z$ are roots of $t^2-2yt+y^2=0,$ which is $(t-y)^2=0.$ Therefore, $x=z=y.$

Now consider the other possibility $y=0$. Then from A.P. of $x,y,z$, $x+z=0 \implies z=-x.$ Also, for the inverse tangents to be in A.P., $2\tan^{-1}0=\tan^{-1}x+\tan^{-1}(-x).$ Since $\tan^{-1}(-x)=-\tan^{-1}x,$ this is true for all $x$. So any triple of the form $(x,0,-x)$ is a valid solution.

4. Check options

We found that the conditions allow:

• $x=y=z$, and also
• $x=-z,\ y=0$.

So the statement "$x=y=z$" is not always necessary.

Now test the options against the general family $(x,0,-x)$:

• A: $x=y=z$ → false in general, e.g. $(1,0,-1)$.
• B: $2x=3y=6z$ → with $(1,0,-1)$ gives $2\neq 0\neq -6$, false.
• C: $6x=3y=2z$ → with $(1,0,-1)$ gives $6\neq 0\neq -2$, false.
• D: $6x=4y=3z$ → with $(1,0,-1)$ gives $6\neq 0\neq -3$, false.

So none of the given options is universally true.

5. Conclusion

The actual solution set is: $x=y=z\quad \text{or} \quad (x,y,z)=(a,0,-a).$

Hence no listed option is correct. In particular, option A alone is not sufficient.