= Solution
By <Fermat's little theorem>,
$$
g(x,y,z)=1-f(x,y,z)^{p-1}\equiv
\begin{cases}
1&\text{if }f(x,y,z)\equiv0\pmod p,\\
0&\text{otherwise}.
\end{cases}
$$
Thus the sum of $g$ over $\mathbb F_p^3$ counts, modulo $p$, the triples on which the <ternary quadratic form> $f$ vanishes.
Expand $f^{p-1}$ with the <multinomial theorem>. A term indexed by $r+s+t=p-1$ contributes a constant multiple of
$$
S_{2r}S_{2s}S_{2t}.
$$
At least one of $r,s,t$ is strictly less than $(p-1)/2$. Its corresponding exponent is therefore in $\{0,1,\ldots,p-2\}$, so part (b), including its $k=0$ case, makes that power sum zero modulo $p$. Every term in the expansion consequently vanishes after summation. The constant term contributes $p^3\equiv0$, and therefore
$$
\boxed{\sum_{x,y,z\in\mathbb F_p}g(x,y,z)\equiv0\pmod p.}
$$
The zero triple is one solution of $f=0$. If it were the only solution, the displayed sum would be congruent to $1$ rather than $0$. Hence there is another triple $(x,y,z)$, not all divisible by $p$, satisfying
$$
\boxed{a x^2+b y^2+c z^2\equiv0\pmod p.}
$$
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