Likely a mistake, but got very large exceptional set in Vojta's more general abc conjecture.
In A more general abc conjecture, p. 7 Paul Vojta conjectures:
If $x_0,\ldots x_{n-1}$ are nonzero coprime integers satisfying $x_0 + \cdots x_{n-1}=0$
$$ \max\{|x_0|,\ldots |x_{n-1}|\} \le C \prod_{p\mid x_0 \cdots x_{n-1}}p^{1+\epsilon}\qquad (1) $$
for all $x_0 , \ldots, x_{n-1}$ as above outside a proper Zariski-closed subset.
Define the quality $q(x_0 \ldots x_{n-1})=\frac{\log(\max(|x_i|))}{\log\operatorname{rad}(x_0 \cdots x_{n-1})}$. The conjecture implies $\limsup q(x_0 \ldots x_{n-1})=1$ outside a proper Zariski-closed subset.
Consider the set of all genus $0$ curves parametrized by two quadratics. $x=a t^2+b t + c,y=d t^2+ e t +f$. Assume $a,b$ are nonzero and $|a|\ge |d$ and $x,y$ are coprime over $\mathbb{Q}[t]$.
The explicit form of the curve is
$$ d^{2} x^{2} + (-2 a d) x y + a^{2} y^{2} + (-2 c d^{2} + b d e - a e^{2} + 2 a d f) x + (- b^{2} d + 2 a c d + a b e - 2 a^{2} f) y + c^{2} d^{2} - b c d e + a c e^{2} + b^{2} d f - 2 a c d f - a b e f + a^{2} f^{2}=0 \qquad(1)$$
Fix $a,b,c,d,e,f$. Take $x_i$ the monomials in $x,y$ of (1), i.e. $x_0 =d^{2} x^{2} ,x_1= (-2 a d) x y$. For infinitely many of them we get quality $4/3-o(1)$ infinitely often for $x,y$ on (1).
$\max|x_i|=a^2 y^2 \sim a^2 d t^4$. The radical is $poly(a,b,c,d,e,f) x y \sim t^4$.
One can make the radical of the quadratic $x$ only $O(t)$ in at least two ways: (a) there exist integers $k$ such that $a t^2+b t + c=k r^2$ has infinitely many integral solutions and (b) find integer $p$ such that for arbitrary large $n, a t^2+b t + c \equiv 0 \pmod{p^n}$ via Hensel lifting. The gcd of $x_i$ is bounded by the resultants and clearing bounded gcd doesn't decrease the quality.
Both (a) and (b) make the radical only $O(t^3)$ for arbitrary large $t$, which gives quality $4/3-o(1)$ and we can get rid of the $o(1)$.
So (1) is in exceptional Vojta's set. By gross abuse of notation, "almost all" $a,b,c,d,e$ give quality $4/3-o(1)$ so we can treat them as variables in (1). This in general makes the dimension of (1) $7$, which is greater than the dimension of $x_0 + \cdots + x_5=0$.
I believe higher dimensional variety can't be subvariety of lower dimensional variety.
Why isn't this construction counterexample?
