= Solution
The parameter relations give $\mu=\log(n/c)$ and
$$
p=\left(\frac{\log(n/c)}{\binom{n-1}{2}}\right)^{1/3}
=\Theta\bigl(n^{-2/3}(\log n)^{1/3}\bigr).
$$
For each fixed integer $r\ge1$, take an $r$-vertex set $S$ and let $X_S$ count triangles meeting $S$. The event that every <vertex> of $S$ is counted by $T$ is exactly $X_S=0$.
The number of such triangles is
$$
N_S=r\binom{n-r}{2}+\binom r2(n-r)+\binom r3
=r\binom{n-1}{2}-\binom r2(n-r)-2\binom r3.
$$
Thus their total expected count is $\mu_S=N_Sp^3=r\mu+O_r(np^3)=r\mu+o(1)$.
Distinct triangles have dependent containment events only when they share an <edge>. If that common <edge> meets $S$, there are $O_r(n)$ choices for it and $O(n^2)$ choices for the two other <vertices>. If the common <edge> avoids $S$, both other <vertices> must lie in $S$, giving only $O_r(n^2)$ pairs. Each pair needs five <edges>. Consequently its ordered <Janson dependency sum> is
$$
\Delta_S=O_r(n^3p^5)
=O_r\bigl(n^{-1/3}(\log n)^{5/3}\bigr)=o(1).
$$
<Janson inequality> gives the upper bound $\Pr(X_S=0)\le\exp(-\mu_S+\Delta_S/2)$. For the lower bound, the avoidance of each triangle is a <decreasing event>. Repeated application of <Harris' inequality> gives
$$
\Pr(X_S=0)\ge(1-p^3)^{N_S}
=\exp[-\mu_S+O_r(n^2p^6)].
$$
The error $n^2p^6=O((\log n)^2/n^2)$ also tends to zero. The bounds match multiplicatively:
$$
\boxed{\Pr(X_S=0)=e^{-r\mu+o(1)}=(c/n)^r(1+o(1)).}
$$
Summing over ordered distinct <vertices> yields the <factorial moments>
$$
\mathbb E(T)_r=(n)_r\Pr(X_S=0)\longrightarrow c^r.
$$
Here is an explicit justification that these moments determine the limit. For every integer $t\ge0$, the finite alternating sums
$$
\frac1{t!}\sum_{j=0}^{L}\frac{(-1)^j}{j!}\mathbb E(T)_{t+j}
$$
are upper bounds for $\Pr(T=t)$ when $L$ is even and lower bounds when $L$ is odd. This follows by applying the <Bonferroni inequalities> to $\binom{T}{t}\sum_j(-1)^j\binom{T-t}{j}$; the full sum is precisely the indicator of $T=t$. First let $n\to\infty$ for a fixed truncation and use the factorial-moment limits. Then let the even and odd truncations grow. Both bounds tend to
$$
\frac{c^t}{t!}\sum_{j=0}^\infty\frac{(-c)^j}{j!}=e^{-c}\frac{c^t}{t!}.
$$
Hence every point <probability> has the <Poisson distribution> limit, proving the <Poisson threshold for vertices lying in no triangle>:
$$
\boxed{T\ \xrightarrow{\ d\ }\ \operatorname{Po}(c).}
$$
This is also the <factorial-moment criterion for Poisson convergence>, here justified rather than left as an unproved implication.
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