= Solution
The ideal $\mathfrak p_2=(2,\alpha)$ has <ideal norm> two. Its square is contained in $(2)$ and has norm four, so $(2)=\mathfrak p_2^2$. A generator would have <field norm> $u^2+26v^2=2$, impossible for integers $u,v$. Thus its <ideal class> has order two.
Modulo three, $X^2+26=(X-1)(X+1)$, giving $(3)=\mathfrak p_3\overline{\mathfrak p}_3$ with $\mathfrak p_3=(3,1-\alpha)$. A generator of $\mathfrak p_3$ would have norm three, again impossible. If $\mathfrak p_3^2$ were principal, its generator would have norm nine. The only possibilities are $\pm3$, whose ideal is $\mathfrak p_3\overline{\mathfrak p}_3$, not $\mathfrak p_3^2$.
Finally $N(1-\alpha)=27$. This element lies in $\mathfrak p_3$ but not $\overline{\mathfrak p}_3$, because $\alpha\equiv-1$ in the latter quotient. Unique <unique factorization of ideals in a number field> and norms therefore give
$$
\boxed{\mathfrak p_3^3=(1-\alpha),\qquad\operatorname{ord}[\mathfrak p_3]=3.}
$$
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