Clearly $\Bbb Z[\sqrt{-6}]$ is not a PID as $\sqrt{-6}$ is irreducible which is not prime. Now we have to find the ideal which is not principal in $\Bbb Z[i\sqrt{6}]$.
I was trying to show that the ideal generated by 3 and $1+\sqrt{-6}$ is not a principal ideal in the ring $\Bbb Z[\sqrt{-6}]$
Suppose that $I=\langle3,1+\sqrt{-6}\rangle$, the ideal in $\mathbb Z[\sqrt{-6}]$ generated by $3$ and $1+\sqrt{-6}$, is principal, say $I=\langle z\rangle$, with $z=a+b\sqrt{-6}$ and $a,b\in\mathbb Z$. The elements $3$ and $1+\sqrt{-6}$ obviously belong to $I$, so they are multiples in $\boldsymbol{\pmb{\mathbb Z}[\sqrt{-6}]}$ of the generator $z$. Therefore we have that $N(z)$ divides both $N(3)=9$ and $N(1+\sqrt{-6})=7$ in $\pmb{\mathbb Z}$, and so $N(z)=1$. In other words, $a^2-6b^2=1$, so necessarily $b=0$ which in turn implies $a^2=1$, so $z=\pm1$, and in particular $\pm z\in I$, so $1\in I$.
Thus, we can write $1=3(a+b\sqrt{-6})+(1+\sqrt{-6})(c+d\sqrt{-6})$, with $c,d,a,b\in\mathbb Z$, which implies $1=3a+c-6d$ and $0=3b+c+d$. Taking classes modulo $3$ we obtain $1=c$ and $0=c+d$, we aren't getting any contradiction shows the result.
I was also trying with $I=\langle5,1+\sqrt{-6}\rangle$ then I got $5a+c-6d=1, 5b+c+d=0$ taking modulo $5$ we didn't get any help.
Can you help me from here?