12. Given $n$ vectors in the plane as $O P_{1}$, $O P_{2}$, $\cdots$, $O P_{n}$, the vector $\boldsymbol{O P}$ makes $\boldsymbol{P P}_{1}^{2}+\boldsymbol{P P}_{2}^{2}+\cdots$ $+P P_{n}^{2}$ minimal. Then the vector $\boldsymbol{O P}$ is $\qquad$
$$
\begin{array}{l}
\text { 12. } \frac{1}{n} \sum_{k=1}^{n} O P_{k} \text {. } \\
\sum_{k=1}^{n} \boldsymbol{P P}_{\boldsymbol{k}}^{2}=\sum_{k=1}^{n}\left(\boldsymbol{O P} \boldsymbol{P}_{\boldsymbol{k}}-\boldsymbol{O P}\right)^{2} \\
=\sum_{k=1}^{n}\left(O P_{k}^{2}-2 O P_{k} \cdot O P+O P^{2}\right) \\
=\sum_{k=1}^{n} O P_{k}^{2}-2\left(\sum_{k=1}^{n} O P_{k}\right) \cdot O P+n O P^{2} \\
=\sum_{k=1}^{n} O P_{k}^{2}-\frac{\left(\sum_{k=1}^{n} O P_{k}\right)^{2}}{n}+ \\
n\left[O P-\frac{1}{n}\left(\sum_{k=1}^{n} O P_{k}\right)\right]^{2} \\
\geqslant \sum_{k=1}^{n} O P_{k}^{2}-\frac{\left(\sum_{k=1}^{n} O P_{k}\right)^{2}}{n} \text {. } \\
\end{array}
$$

When $O P=\frac{1}{n} \sum_{k=1}^{n} O P_{k}$, the equality holds in the above expression, hence
$$
O P=\frac{1}{n} \sum_{k=1}^{n} O P_{k}
$$