If following actions allowed, Find $F(2002,2020,2200)?$ If following actions allowed,Find $F(2002,2020,2200)?$
$$ F(x+t,y+t,z+t)=t+F(x,y,z);$$
$$ F(xt,yt,zt)=tF(x,y,z);$$
$$ F(x,y,z)=F(y,x,z)=F(x,z,y)$$ 
where x,y,z,t are real numbers.
My attempt:
$F(0,0,0)=0$ from second action, then $F(x,x,x)=x$ for any x. And $F(x,y,z)=F(x,z,y)=F(y,x,z)=F(y,z,x)=F(z,x,y)=F(z,y,x)$ from third action. 
And I've found $$F(2x,2y,x+y)=2x+F(0,2y-2x,y-x)=2x+(y-x)F(0,2,1)=2y+F(2x-2y,0,x-y)=2y+(x-y)F(2,0,1)$$
Then $2x+(y-x)F(0,1,2)=2y+(x-y)F(0,1,2)$, $F(0,1,2)=1.$
And I'm trying to find $F(0,1,11)$
I need some hint, I can't do more.
In my opinion $F(x,y,z)=(x+y+z)/3$
 A: Let $$g(t)=F(0,1,t) $$
Then $$g(1-t)=F(0,1,1-t)=1+F(-1,0,-t)=1-F(0,1,t)=1-g(t) $$
and for $t\ne0$,
$$g(\tfrac1t)=F(0,1,\tfrac1t)=\tfrac1tF(0,t,1)=\tfrac1tg(t) .$$
Conversely, let $g\colon\Bbb R\to\Bbb R$ be any function with
$$\tag{1}\begin{align}g(1-t)&=1-g(t)\\g(\tfrac1t)&=\frac1tg(t)&(t\ne0)\end{align} $$
and define
$$\tag2F(x,y,z)=\begin{cases}x+(y-x)g(\tfrac{z-x}{y-x})&x\ne y\\
x+(z-x)g(0)&x=y .\end{cases}$$
Then we find
$$F(x+t,y+t,z+t)=t+F(x,y,z) $$
as well as (for $t=0$ trivially, for $t\ne0$ from $(2)$)
$$F(tx,ty,tz)=tF(x,y,z).$$
For $x\ne y$ (and trivially also for $x=y$), 
$$\begin{align}F(y,x,z)&=y+(x-y)g(\tfrac{z-y}{x-y})
\\&=y+(x-y)g(1-\tfrac{z-x}{y-x})
\\&=y+(x-y)(1-g(\tfrac{z-x}{y-x})
)
\\&=x+(y-x)g(\tfrac{z-x}{y-x})
\\&=F(x,y,z). \end{align}$$
If $z\ne x\ne y$, 
$$ 
F(x,z,y)=x+(z-x)g(\tfrac{y-x}{z-x})
=x+(z-x)g(\tfrac1{\tfrac{z-x}{y-x}})
=x+(z-x)\frac{y-x}{z-x}g(\tfrac{z-x}{y-x})
=x+(y-x)g(\tfrac{z-x}{y-x})
=F(x,y,z).$$
If $z=x\ne y$, 
$$ 
F(x,z,y)=x+(y-x)g(0)
=x+(y-x)g(\tfrac{z-x}{y-x})
=F(x,y,z).$$
By symmetry, we also have $F(x,z,y)=F(x,y,z)$ when $z\ne x=y$, and trivially also when $x=y=z$. Thus
$$ F(x,y,z)=F(y,x,z)=F(x,z,y)$$
for all $x,y,z$.
Thus the problem is equivalent to considering functions $g$ with the two properties $(1)$ and to compute $g(11)$.
For $x,y\in\Bbb R$, say $x\sim y$ if we can get from $x$ to $y$  by applying finitely many steps $t\mapsto 1-t$ or (for $t\ne 0$) $t\mapsto \frac1t$. As both allowed steps are involutions, we may assume that the two types of steps are always used alternatingly in such a sequence.
If we start with $x\ne0,1$, we return to $x$ after at most six steps:
$$ x\mapsto  1-x\mapsto \frac1{1-x}\mapsto \frac x{x-1}\mapsto \frac{x-1}{x}\mapsto \frac1x\mapsto x$$
where all intermediate values are also $\ne0,1$
(If we start with $t\mapsto \frac 1t$ instead of $t\mapsto 1-t$, we arrive at the same sequence, just backwards).
We conclude that $\sim$ is an equivalence relation on $\Bbb R\setminus\{0,1\}$.
If we agree that $\frac10=\infty$ and $\frac1\infty=0$, we see that $\sim$ is an equivalence relation on $\Bbb R\cup\{\infty\}$.
For most $t\in\Bbb R\cup\{\infty\}$, the equivalence class $[t]$ has $6$ elements.
As just seen $[0]=[1]=[\infty]$ is an equivalence class with $3$ elements. 
For other $t$, $[t]$ has $3$ elements only if 
$$ t=1-\frac{1-t}\quad \text{or}\quad t=\frac1{1-\frac1t}.$$
We find that this is the case only for $[2]=\{2,\frac12,-1\}$.
$[t]$ has two elements only if 
$$ 1-t=\frac1t,$$
which has no real solutions.
Finally, $[t]$ cannot have only one element as that would need $t=1-t=\frac1t$.
This allows us to describe the set of possible function $g$.


*

*For each $[t]$ of size $6$, we can pick $y=g(t)$ arbitrarily, and use $(1)$ to define $g(1-t)=1-y$, $g(\frac1{1-t})=\frac{1-y}{1-t}$, $g(\frac t{t-1})=\frac{y-t}{1-t}$, $g(\frac{t-1}t)=\frac{t-y}t$, $g(\frac1t)=\frac yt$. This works out becasue in the end we have $g(\frac1t)=\frac1tg(t)$ as desired.

*For $[3]$, define $g(-1)=0$, $g(2)=1$, $g(\frac12)=\frac12$. (As you already showed, these values cannot be played with)

*For $[0]$, defined $g(0)$ arbitrarily and $g(1)=1-g(0)$
Any such $g$ solves the problem. In particular, $g(11)$ can take an arbitrary value and so can
$$F(2002,2020,2200)=2002+18g(11). $$
A: There is a good reason why you can't do more : the problem is under-determined, i.e. there are not enough constraints to force the value of $F(2002,2020,2200)$.
To see why, let $k$ be any constant, and consider the following function.
$$
G(x,y,z)=\left\lbrace\begin{array}{lcl}
x+k(z-x), & \textrm{if} & z-x=\frac{y-x}{11} \\
z, & \textrm{if} & z-x=\frac{y-x}{2} \\
y, & \textrm{if} & z-x=2(y-x) \\
x+k(y-x), & \textrm{if} & z-x=11(y-x) \\
x, & \textrm{otherwise.} \\
\end{array}\right.
$$
By construction, $G$ satisfies $G(t+x,t+y,t+z)=t+G(x,y,z)$ and
$G(tx,ty,tz)=tG(x,y,z)$ for any $t$. Because of the symmetry of $y$ and $z$ in the definition of $G$, we also have $G(x,y,z)=G(x,z,y)$ (although we do not have $G(x,y,z)=G(y,x,z)$). We can now define :
$$
F(x,y,z)=\left\lbrace\begin{array}{lcl}
G(x,y,z), & \textrm{if} & \min(x,y,z) = x \\
G(y,z,x), & \textrm{if} & \min(x,y,z) = y \\
G(z,x,y), & \textrm{if} & \min(x,y,z) = z \\
\end{array}\right.
$$
It is then easy to check, using what we have just remarked on $G$, that $F$ satisfies all your requirements. And
$$
F(2002,2020,2200)=G(2002,2020,2200)=2002+18k
$$
(since here, we have $z-x=198$ and $y-x=18$, so $z-x=11(y-x)$)
So $F(2002,2020,2200)$ can be an arbitrary value.
