An identity associated with the centroid of a triangle 
Four year ago, I am looking for a proof of my identity as follows:

Let $ABC$ be a triangle, let $G$ be the centroid of $ABC$. Let $P$ be any point on the plane. Let $H, N, O$ on the plane such that: $\overrightarrow{HN}:\overrightarrow{NG}:\overrightarrow{GO}=3:1:2$ , then: 
$$PA^2+PB^2+PC^2+PH^2-4PN^2=OA^2+OB^2+OC^2$$
Four special case:
1) When $O$ is the circumcenter, $H$ is the orthocenter, $N$ is Nine point center:
$$PA^2+PB^2+PC^2+PH^2-4PN^2=3R^2$$
2) When $O$ is the circumcenter, $H$ is the orthocenter, $N$ is Nine point center, and $P$ lie on Nine point center.
$$PA^2+PB^2+PC^2+PH^2=4R^2$$
3) when $G \equiv O \equiv H \equiv N$ then $$PA^2+PB^2+PC^2=GA^2+GB^2+GC^2+3PG^2$$
4) when  $O$ is the circumcenter and $P$ is the Nine point center then 
$$NA^2+NB^2+NC^2+NH^2=3R^2$$

Michael's proof is more elegant, but it is also possible to hack it out with vectors.
Let $G$ be the origin of a cartesian coordinate system, and let boldface lowercase letter denotes the position of the uppercase letter.  Then $\mathbf{a}+\mathbf{b}+\mathbf{c}=\vec{0}$, $\mathbf{h}=4\mathbf{n}=-2\mathbf{o}$, so
\begin{align*}\require{cancel}
LHS&=(\mathbf{p}-\mathbf{a})^2+(\mathbf{p}-\mathbf{b})^2+(\mathbf{p}-\mathbf{c})^2+(\mathbf{p}-\mathbf{h})^2-4(\mathbf{p}-\mathbf{n})^2\\
&=2\mathbf{p}\cdot(4\mathbf{n}-\mathbf{a}-\mathbf{b}-\mathbf{c}-\mathbf{h})+\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2+\mathbf{h}^2-4\mathbf{n}^2\\
&=\cancel{2\mathbf{p}\cdot\vec{0}}+\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2+12\mathbf{n}^2\\
RHS &=(\mathbf{a}-\mathbf{o})^2+(\mathbf{b}-\mathbf{o})^2+(\mathbf{c}-\mathbf{o})^2\\
&=\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2-\cancel{2\mathbf{o}\cdot(\mathbf{a}+\mathbf{b}+\mathbf{c})}+3\mathbf{o}^2\\
&=\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2+3\mathbf{o}^2\\
&=\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2+12\mathbf{n}^2.
\end{align*}
