= Solution
The <Hasse-Minkowski theorem> says that a nondegenerate <quadratic form> over a <number field> has a nonzero isotropic vector over that field if and only if it has one over every completion. In particular, over $\mathbb Q$ one tests $\mathbb R$ and every $\mathbb Q_\ell$. For the present integral form, a rational isotropic vector can be multiplied by a common denominator to give an integer one. This is the <Hasse-Minkowski principle>.
Write $P=p$, $R=p+12$ and $S=p+24$. All three primes are odd and exceed three: $p=2$ or $3$ would make $p+12$ composite. The <second supplementary law for quadratic reciprocity> gives
$$
\left(\frac2P\right)=(-1)^{(P^2-1)/8}.
$$
\b[Necessity at $P$.] Scale any nonzero vector over $\mathbb Q_P$ so that its coordinates lie in $\mathbb Z_P$ and at least one is a unit. Reduction of its equation modulo $P$ gives $y^2=2z^2$, since twelve is invertible modulo $P$. If the <Legendre symbol> $(2/P)$ is $-1$, this forces $P\mid y,z$. The original equation then forces $P\mid x$, because its other two terms are divisible by $P^2$. This contradicts the normalization. Thus
$$
\boxed{\text{a nonzero solution requires }p\equiv1\text{ or }7\pmod8.}
$$
\b[Sufficiency at the coefficient primes.] Suppose $(2/P)=1$. A nonzero solution modulo $P$ is $(x,y,z)=(0,\sqrt2,1)$. Its $y$-derivative is nonzero modulo $P$, so fixing the other two coordinates and applying the <Hensel lemma> lifts it to $\mathbb Q_P$.
Since $S\equiv P\pmod8$, the same <Legendre symbol> criterion gives $(2/S)=1$. Modulo $S$, the equation reduces to $y^2=2x^2$, and the vector $(1,\sqrt2,0)$ is a simple zero in the $y$ coordinate. It lifts to $\mathbb Q_S$. Modulo $R$, the coefficients of $x^2$ and $z^2$ are respectively $-12$ and $12$, so $(1,0,1)$ is a zero with nonzero $x$-derivative. It lifts to $\mathbb Q_R$.
\b[The remaining odd primes.] If $\ell$ is odd and divides none of $PRS$, the reduced <ternary quadratic form> is nondegenerate. An elementary finite-field argument finds a zero with $z=1$: if its coefficients are $A,B,C$, the sets
$$
\{A x^2:x\in\mathbb F_\ell\},\qquad
\{-C-B y^2:y\in\mathbb F_\ell\}
$$
both have $(\ell+1)/2$ elements, so intersect. The resulting vector has nonzero $z$-derivative $2C$, and the <Hensel lemma> lifts it. This <isotropy of nondegenerate ternary quadratic forms over finite fields> verifies every remaining odd completion without an extra congruence condition.
\b[The dyadic and real places.] Setting $y=1,z=2$ reduces the dyadic problem to
$$
x^2=-3-\frac{84}{P}.
$$
The right side is a <2-adic unit> congruent to $-3-4=1$ modulo eight. The <unit square classes of the p-adic integers> computed above show that it is a square in $\mathbb Q_2$. This also explains the printed hint: its integer can be taken as $m=-(p+21)/2$. Over $\mathbb R$, the positive and negative coefficients make the form indefinite; for instance $x=0,z=1,y=\sqrt{S/R}$ gives a nonzero zero.
All completions now have a nonzero isotropic vector. The <Hasse-Minkowski theorem>, followed by clearing denominators, proves
$$
\boxed{\text{a nonzero integer solution exists }\iff p\equiv\pm1\pmod8.}
$$
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