, for some $i\in\{1,2\}$, we have that $|\cA(i)|\le |\cA(1,2)|$ then $(\cI\cup\{A\setminus \{i\}\mid A\in\cA(1,2)\}\cup\{\{1,2\}\setminus\{i\}\})\setminus\cA(i)$ is a star-subfamily of $\cH$ of size larger than $\cI$, which is a contradiction. Thus we know that $|\cA(1,2)|<\min(|\cA(1)|,|\cA(2)|)$.\
If $|\cA(1)|=|\cA(2)|=1$, then $\cA(1,2)=\emptyset$, and $(\cI\cup\{\{1,j\}:j\in C(1)\}\cup\{\{1\}\})\setminus\cA(2)$ is a star subfamily of size larger that $\cI$, a contradiction., so we may assume without loss of generality that $|\cA(1)|\ge 2$.\
For $i\in\{1,2\}$, set $\cA^\pr(i)=\{A\setminus\{i\}\mid A\in \cA(i)\}$; then $|\cA^\pr(i)|=|\cA(i)|$. Clearly $\cA^\pr(1)$ and $\cA^\pr(2)$ cross-intersect. If, for some $i\in\{1,2\}$, $\cA^\pr(i)$ is an intersecting family then $\cI\cup\cA^\pr(i)\setminus\cA(1,2)$ is an intersecting subfamily of $\cH$ that is larger that $\cI$, a contradiction, so we have that neither $\cA^\pr(1)$ nor $\cA^\pr(2)$ is intersecting. Since $\cA^\pr(1)$ is not intersecting and $|\cA^\pr(1)|\ge 2$, we may assume (by relabeling, if necessary) that $\{\{3,4\},\{5,6\}\}\subseteq\cA^\pr(1)$. Because $\cA^\pr(2)$ cross-intersects $\cA^\pr(1)$ we have $\cA^\pr(2)\subseteq\{\{3,5\},\{3,6\},\{4,5\},\{4,6\}\}$. In particular, $|\cA(2)|=|\cA^\pr(2)|\le 4$ and, for each $x\in\{3,4,5,6\}$, $\{1,x\}$ is a subset of some set in $\cA(1)$. But then $(\cI\setminus\cA(2)) \cup \{\{1,x\}\mid x\in\{3,4,5,6\}\}\cup\{\{1\}\}$ is an intersecting subfamily of $\cH$ that is larger than $\cI$, a contradiction.\

$|\cIt|=0$
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Here $\cIt=\emptyset$ and $\cI$ is an intersecting family of $3$-sets such that no 2-subset of $[n]$ is contained in every element of $\cI$ (otherwise that $2$-subset could be added to $\cI$).\
Let $\cS$ be the largest star in $\cI$ (clearly $|\cS|\ge 2$), and let $D$ be the head of $\cS$. If $\cS=\cI$ then we are done, so define $\cR=\cI\setminus\cS$ and assume that $\cR\ne\emptyset$. In particular, for every $R\in\cR$ we must have that $R\cap D=\emptyset$; otherwise $R$ could be added 