Call (the projective completion of) your surface $S$. It admits three double covers of $\mathbb P^2$, namely $$\pi_1(a,b,c)=(a,b), \quad\pi_2(a,b,c)=(a,c),\quad\pi_3(a,b,c)=(b,c).$$ Each double cover induces an involution, so we get three involutions $$\sigma_1,\sigma_2\sigma,\sigma_3:S\to S.$$$$\sigma_1,\sigma_2,\sigma_3:S\to S.$$ These involutations don't commute, and if you form $f=\sigma_1\circ\sigma_2$ and $g=\sigma_1\circ\sigma_3$, then $f$ and $g$ generate a subgroup $G$ of $\text{Aut}(S)$ that is a free group on two generators. And presumably for most starting points $P\in S(\mathbb{Q})$, repeated application of $f$ and $g$ will give you a tree of rational solutions. (You might compare this with the generation of all positive integer solutions to the Hurwitz equation $x^2+y^2+z^2=3xyz$ starting from $(1,1,1)$.) Here are some references to papers that have studied rational points on K3 surfaces admitting 3 involutions:
Baragar, A. Rational points on K3 surfaces in $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$. Math. Ann. 305 (1996), no. 3, 541–558.
Wang, L., Rational Points and Canonical Heights on K3-surfaces in $\mathbb P^1\times\mathbb P^1\times\mathbb P^1$, Contemporary Math. 186 (1995), 273 – 289
