(Edit at the bottom.) Here is an elementary way (known to Fermat) to find an infinite number of rational points. From $a+b+c = abc = 6$, we need to solve the equation,
$$ab(6-a-b) = 6\tag1$$
Solving $(1)$ as a quadratic in $b$, its discriminant $D$ must be made a square,
$$D := a^4-12a^3+36a^2-24a = z^2$$
Using any non-zero solution $a_0$, do the transformation,
$$a=x+a_0\tag2$$
For this curve, let $a_0=2$, and we get,
$$x^4-4x^3-12x^2+8x+16$$
Assume it to be a square,
$$x^4-4x^3-12x^2+8x+16 = (px^2+qx+r)^2$$
Expand, then collect powers of $x$ to get the form,
$$p_4x^4+p_3x^3+p_2x^2+p_1x+p_0 = 0$$
where the $p_i$ are polynomials in $p,q,r$. Then solve the system of three equations $p_2 = p_1 = p_0 = 0$ using the three unknowns $p,q,r$. One ends up with,
$$105/64x^4+3/8x^3=0$$
Thus, $x =-16/35$ or,
$$a = x+a_0 = -16/35+2 = 54/35$$
and you have a new rational point,
$$a_1 = 54/35 = 6\times 3^{\color{red}2}/35$$
Use this on $(2)$ as $x = y+54/35$ and repeat the procedure. One gets,
$$a_2 = 6\times 4286835^{\color{red}2}/37065988023371$$
Again using this on $(2)$, we eventually have,
$$\small {a_3 = 6\times 11838631447160215184123872719289314446636565357654770746958595}^{\color{red}2} /d\quad$$
where the denominator $d$ is a large integer too tedious to write.
Conclusion: Starting with a "seed" solution, just a few iterations of this procedure has yielded $a_i$ with a similar form $6n^{\color{red}2}/d$ that grow rapidly in "height". Heuristically, it then suggests an infinite sequence of distinct rational $a_i$ that grow in height with each iteration.
$\color{blue}{Edit}$: Courtesy of Aretino's remark below, then another piece of the puzzle was found. We can translate his recursion into an identity. If,
$$a^4-12a^3+36a^2-24a = z^2$$
then subsequent ones are,
$$v^4-12v^3+36v^2-24v = \left(\frac{12\,e\,g\,(e^2+3f^2)}{(e^2-f^2)^2}\right)^2$$
where,
$$\begin{aligned}
v &=\frac{-6g^2}{e^2-f^2}\\
\text{and,}\\
e &=\frac{a^3-3a^2+3}{3a}\\
f &=\frac{a^3-6a^2+9a-6}{z}\\
g &=\frac{a^3-6a^2+12a-6}{z}
\end{aligned}$$
Starting with $a_0=2$, this leads to $v_1 = 6\times 3^2/35$, then $v_2 = 6\times 4286835^2/37065988023371$, ad infinitum. Thus, this is an elementary demonstration that there an infinite sequence of rational $a_i = v_i$ without appealing to elliptic curves.
Best Answer
Given any rational $$ u^2 + v^2 = 1,$$ we find $$ (u-v)^2 + (u+v)^2 = 2 $$
Meanwhile, given any integer Pythagorean triple $$ a^2 + b^2 = c^2, $$ we get rational $$ \left( \frac{a}{c} \right)^2 + \left( \frac{b}{c} \right)^2 = 1 $$
From the comments above, I guess I should add that all primitive solutions in integers to $x^2 + y^2 = 2 z^2,$ since this means $x \equiv y \pmod 2$ and $z$ odd, so we get $$ \left( \frac{x-y}{2} \right)^2 + \left( \frac{x+y}{2} \right)^2 = z^2 $$ in integers and primitive. We can find all of these with the tandard parametrization for primitive Pythagorean triples.