5. $\vec{a}=\left(k^{2}+k-3\right) \vec{i}+(1-k) \vec{j}, \vec{b}=-3 \vec{i}+(k-1) \vec{j}$, vectors $\vec{a}$ and $\vec{b}$ are collinear, $k=$ $\qquad$
5. $-3,2,1$. $\left(k^{2}+k-3\right)(k-1)=-3(1-k)$.

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

5. $-3,2,1$. $\left(k^{2}+k-3\right)(k-1)=-3(1-k)$.