= Solution
Because $p\equiv11\pmod{12}$, we have $p\equiv3\pmod4$ and
$$
\mathcal O_K=\mathbb Z\left[\frac{1+\sqrt{-p}}2\right].
$$
For $\theta=(1+\sqrt{-p})/2$, the <minimal polynomial> is
$$
f(T)=T^2-T+\frac{p+1}{4}.
$$
Since $p\equiv2\pmod3$, reduction modulo $3$ gives $f(T)\equiv T(T-1)$. Its two roots are distinct, so
$$
(3)=\mathfrak p_1\mathfrak p_2
$$
with distinct prime ideals of norm $3$. Thus $3$ splits completely, as recorded in <splitting of three in Q of square root minus p>.
The <ideal class group> $\operatorname{Cl}_K$ is the group of nonzero fractional ideals of $\mathcal O_K$ modulo the subgroup of principal fractional ideals. Saying that $\mathfrak p_1$ has order $n$ means that $n$ is the least positive integer for which $\mathfrak p_1^n$ is principal.
Suppose for contradiction that $n$ is odd and $\mathfrak p_1$ has order $n$. Then $\mathfrak p_1^n=(\alpha)$ for some $\alpha\in\mathcal O_K$, and taking <ideal norm>[norms] gives
$$
|N_{K/\mathbb Q}(\alpha)|=N(\mathfrak p_1)^n=3^n.
$$
Write $\alpha=(a+b\sqrt{-p})/2$, where $a,b\in\mathbb Z$ have the same parity. Then
$$
4\cdot3^n=a^2+pb^2.
$$
If $b\ne0$, the right-hand side is at least $p$, contradicting $p>3^{n+2}>4\cdot3^n$. Hence $b=0$, so $\alpha$ is a rational algebraic integer and therefore an integer. But then $3^n=N(\alpha)=\alpha^2$, impossible when $n$ is odd. This proves the <odd-order obstruction for a split prime in an imaginary quadratic field> and shows that the order cannot equal $n$.
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