Finding a specific function on $\mathbb{R}^3$ with some symmetry properties and partial knowledge of the function

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I've been stumped with a problem for some time and thought I'd try my chance here. I am looking for a function $\vec{a}:\mathbb{R}_+^3\to \mathbb{R}_+^3$ such that if $\vec{a}=(a_1,a_2,a_3)$, then the following properties hold:

  1. $a_1, a_2, a_3$ are defined on $\{(x,y,z):0\leq x+y+z\leq 1\}\cap \mathbb{R}^3_+$

  2. For any $(x,y,z)$ in the domain, $a_1(x,y,z)+a_2(x,y,z)+a_3(x,y,z)=1$.

  3. $a_1(x,y,z)=a_1(x,z,y), \quad a_2(x,y,z)=a_2(z,y,x), \quad a_3(x,y,z)=a_3(y,x,z).$

  4. $a_1(x,y,z)=a_2(y,x,z)=a_3(y,z,x).$

  5. If $x=y$, then $$a_1(x,y,z)=\frac{2}{3}-\frac{1}{3}\sqrt{1-3x^2+3z^2}.$$ I'd be happy to see any function that satisfies all these properties, although what I'm after is a specific function about which the only additional thing I can say is that I suspect it's probably as simple as it can be, given the 5 properties. Any hints on how even to approach this effectively would be appreciated!