Eliminating $r$ from $6\tan(r+x)=3\tan(r+y)=2\tan(r+z)$ Here is the question:
If $$\boldsymbol{6\tan (r+x)=3\tan(r+y)=2\tan(r+z)}$$
show that
$$\boldsymbol{3\sin^2(x-y)+5\sin^2(y-z)-2\sin^2(z-x)=0}$$
It seems easy but the natural approach goes off the rails.
What I have so far,
Take the two equalities and cross multiply,
$$2\sin(r+x)\cos(r+y)=\sin(r+y)\cos(r+x)$$
$$3\sin(r+y)\cos(r+z)=2\sin (r+z)\cos(r+y)$$
then use trigonometric addition formula,
$$2\sin(2r+x+y)+2\sin(x-y)=\sin (2r+x+y)+\sin(y-x)$$
$$3\sin(2r+y+z)+3\sin(y-z)=2\sin(2r+y+z)+2\sin(z-y)$$
And get
$$\sin(2r+x+y)=3\sin(y-x)$$
$$\sin(2r+y+z)=5\sin(z-y)$$
Now use the sum formula for sine and solve the linear system giving,
$$\sin 2r \sin(x-z)=3\sin(y-x)\sin(y+z)-5\sin(z-y)\sin(x+y)$$
$$\cos 2r \sin (x-z) =3\sin(y-x)\cos(y+z)-5\sin(z-y)\cos(x+y)$$
If we square them and add we eliminate $r$. and get
$$\sin^2 (x-z)=9\sin^2(y-x)+25\sin^2(z-y)$$
$$-30\sin(y-x)\sin(z-y)\cos(x-z)$$
But how to go from here ? This doesnt look anything like the desired formula.
Can there be an error in the original problem ? Am I making an error ? Or missing something ?

Just as you derive,
$$\sin(2r+x+y)=3\sin(y-x)\tag1$$
$$\sin(2r+y+z)=5\sin(z-y)\tag2$$
you can get, $$\sin(2r+z+x)=-2\sin(z-x)\tag3$$
From (1), can you get, $$\begin{align}\sin(2r+x+y)\sin(y-x)&=3\sin^2(y-x)\\ \frac12\left[\cos(2r+2x)-\cos(2r+2y)\right]&=3\sin^2(y-x)\tag{1´}\end{align}$$?
From (2), can you get, $$\begin{align} \sin(2r+y+z)\sin(z-y)&=5\sin^2(z-y)\\ \frac12\left[\cos(2r+2y)-\cos(2r+2z)\right]&=5\sin^2(z-y)\tag{2´}\end{align}$$?
Similarly, from (3), can you get, $$\begin{align} \sin(2r+z+x)\sin(x-z)&=-2\sin^2(z-x)\\ \frac12\left[\cos(2r+2z)-\cos(2r+2x)\right]&=-2\sin^2(z-x)\tag{3´}\end{align}$$?
Observe what happens when you add, $(1´)$, $(2´)$ and $(3´)$.

Additionally, you can solve this more general problem:

If $$\small \boldsymbol{a\tan (r+x)=b\tan(r+y)=c\tan(r+z)}$$
show that $$\small \boldsymbol{\left(\frac{a+b}{a-b}\right)\sin^2(x-y)+\left(\frac{b+c}{b-c}\right)\sin^2(y-z)+\left(\frac{c+a}{c-a}\right)\sin^2(z-x)=0}.$$

and enjoy! : )
