On integers as sums of three integer cubes revisited

A parametrization with quadratic forms like you want defines a conic inside the projective cubic surface with equation (3). Such a conic is contained in a plane and the residual intersection of the plane with the surface is a line. Conversely any plane containing a line inside the cubic surface will have as residual intersection a conic that can be parametrized by quadratic forms. A smooth cubic surface such as (3) famously has 27 lines. If you want your quadratics to have rational coefficients, since you are interested in integral solutions,you need the line to be defined over the rationals. So you need to go through the 27 lines and see which are rational. The ones easy to eyeball ($x_i = \omega x_j, x_k = \alpha x_4, \omega^3=-1,\alpha^3 = N$ with $i,j,k$ a permutation of $1,2,3$. Did I get 27?) aren't defined over the rationals, so you are out of luck.

Perhaps if you start with my three-rational-cubes identity $$ ab^2 = \biggl(\frac{(a^2+3b^2)^3+(a^2-3b^2)(6ab)^2}{6a(a^2+3b^2)^2}\biggr)^{\!3} - \biggl(\frac{(a^2+3b^2)^2-(6ab)^2}{6a(a^2+3b^2)}\biggr)^{\!3} - \biggl(\frac{(a^2-3b^2)6ab^2}{(a^2+3b^2)^2}\biggr)^{\!3} $$ you might be able to find something?