10.2 .5 * Find a point $P$ inside $\triangle ABC$ such that $AP^{2}+BP^{2}+$ $CP^{2}$ is minimized.
Let $\overrightarrow{C A}=\alpha, \overrightarrow{C B}=\boldsymbol{\beta}, \overrightarrow{C P}=p$, then $\overrightarrow{A P}=p- \boldsymbol{\alpha}, \overrightarrow{B P}=\boldsymbol{p}-\boldsymbol{\beta}, A P^{2}+B P^{2}+C P^{2}=(\boldsymbol{p}-\boldsymbol{\alpha})^{2}+(\boldsymbol{p}- \boldsymbol{\beta})^{2}+\boldsymbol{p}^{2}=3 \boldsymbol{p}^{2}-2 \boldsymbol{p} \cdot(\boldsymbol{\alpha}+\boldsymbol{\beta})+\boldsymbol{\alpha}^{2}+\boldsymbol{\beta}^{2}=3\left[\boldsymbol{p}-\frac{1}{3}(\boldsymbol{\alpha}+ \boldsymbol{\beta})\right]^{2}+\boldsymbol{\alpha}^{2}+\boldsymbol{\beta}^{2}-\frac{1}{3}(\boldsymbol{\alpha}+\boldsymbol{\beta})^{2}$. From this, we can conclude that when $\boldsymbol{p}=\frac{1}{3}(\boldsymbol{\alpha}+ \beta)$, the sum of these squares is minimized. At this point, the point $P$ is precisely the centroid of $\triangle A B C$.