jmc

Let $z=3x+4y.$ Then $y=\frac{z-3x}{4}.$ Substituting into $x^2+y^2=14x+6y+6,$ we get $$x^2+{\left(\frac{z-3x}{4}\right)}^2=14x+6\cdot \frac{z-3x}{4}+6.$$This simplifies to $$25x^2-6xz+z^2-152x-24z-96=0.$$Writing this as a quadratic in $x,$ we get $$25x^2-(6z+152)x+z^2-24z-96=0.$$This quadratic has real roots, so its discriminant is nonnnegative. This gives us $$(6z+152{)}^2-4\cdot 25\cdot (z^2-24z-96)\ge 0.$$This simplifies to $-64z^2+4224z+32704\ge 0,$ which factors as $-64(z+7)(z-73)\ge 0.$ Therefore, $z\le 73.$

Equality occurs when $x=\frac{59}{5}$ and $y=\frac{47}{5},$ so the maximum value is $\boxed{73}.$