<s>[INST] How to get the $INR(x_i)$ in PCA, the relative contribution of $x_i$ to the total inertia? Given the following data array :
$$\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
J/I&1 & 2 & 3 & 4 & 5 & 6\\
\hline
x & 1 & 0 & 0 & 2 & 1 & 2\\
y & 0 & 0 & 1 & 2 & 0 & 3\\
z & 0 & 1 & 2 & 1 & 0 & 2\\
\hline
\end{array}$$
I can get the following values for the centered data $Y$ along with the variance
$$\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
&1 & 2 & 3 & 4 & 5 & 6 & v_2 & v_3\\
\hline
\mbox{x }&1 & 0 & 0 & 2 & 1 & 2 & 1/\sqrt{6} & 1/\sqrt{2}\\
\mbox{y }&0 & 0 & 1 & 2 & 0 & 3 & 2/\sqrt{6} & 0\\
\mbox{z }&0 & 1 & 2 & 1 & 0 & 2 & 1/\sqrt{6} & -1/\sqrt{2}\\
\hline
\end{array}$$
From there I can get the principal components value :
\begin{align}
Vv_2&=
\begin{pmatrix}4 & 4 & 0\\
4 & 8 & 4\\
0 & 4 & 4\end{pmatrix}
\frac{1}{\sqrt 6}\begin{pmatrix}
1\\2\\1
\end{pmatrix}\\
&=\frac{2}{\sqrt 6}
\begin{pmatrix}
1\\2\\1
\end{pmatrix}
\end{align}
$PC_1$ value is therefore $2$. 
\begin{align}
Vv_3&=
\begin{pmatrix}4 & 4 & 0\\
4 & 8 & 4\\
0 & 4 & 4\end{pmatrix}
\frac{1}{\sqrt 2}\begin{pmatrix}
1\\0\\-1
\end{pmatrix}\\
&=\frac{2}{3\sqrt 2}
\begin{pmatrix}
1\\0\\-1
\end{pmatrix}
\end{align}
$PC_1$ value is therefore $\frac{2}{3}$
$$\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
&1 & 2 & 3 & 4 & 5 & 6 & v_2 & v_3\\
\hline
\mbox{value $PC_1$ }&&&&&&&2&\\
\mbox{Value $PC_2$ }&&&&&&&&2/3\\
\mbox{value coef INR }&&&&&&&&\\
\mbox{value coef CTR }&&&&&&&&\\
\mbox{value coef COR }&&&&&&&&\\
\hline
\end{array}$$
How to get the INR (I think it's a French acronym) the contribution of an individual $x_i$ to the total inertia $I_T$:
$$INR(i)=\frac{p_id(0,y_i)^2}{I_T}$$
With $d$ being usually the euclidean distance. We can deduce from the definition that $\sum INR(i)=1$.
\begin{align}
INR(1) &= \frac{1}{6}\times\frac{(-1)²+1²}{2+\frac{2}{3}} = \frac{1}{4}\\
INR(2) &= \frac{1}{6}\times\frac{(-1)²+(-1)²}{2+\frac{2}{3}} = \frac{1}{4}\\
INR(3) &= \frac{1}{6}\times\frac{1²+1²}{2+\frac{2}{3}} = \frac{1}{4}\\
INR(4) &= \frac{1}{6}\times\frac{1²+2²+1²}{2+\frac{2}{3}} = \frac{2}{3}???\\
\end{align}
It would be much more than $1$ now.
 [/INST] If I'm reading your question correctly, you're looking for inertia around an arbitrary point in your data cloud. The can be formulated as: $I_g-||\bar x-a||^2$.  where.  $I_g$ here is total inertia, $a$ is particular point in question.  
Here's a source on that: with derivation included. Page 8-9.
https://cedric.cnam.fr/fichiers/art_1827.pdf
</s>