Finding center of mass for tetrahedron I am given a tetrahedron with the following points: 
$$\begin{align}
P_1 &= (2,0,1)\\
P_2 &= (-1,1,1)\\
P_3 &= (1,0,2)\\
P_4 &= (3,1,4)
\end{align}$$
and I am tasked with finding its center of mass, M.
My attempt: 
$$\vec{OM} = \frac{1}{3}(\vec{OP_1} + \vec{OP_2} + \vec{OP_3})$$
$\vec{OM_1}$ is the coordinate for the center of mass of the base and $\vec{OM_2}$ is the coordinate of the center of mass of the tetrahedron. To find the center of mass of the tetrahedron, I do the following: 
$$\begin{align}
\vec{OM_2} &= \frac{\vec{OM} - \vec{OP_4}}{2} \\
\vec{OM_2} &= \frac{\frac{1}{3}(2,1,4) - (3,1,4)}{2} \\
\vec{OM_2} &= (-\frac{1}{3}, -\frac{1}{6}, -\frac{8}{6})
\end{align}$$
This is wrong since the answer is: $$(\frac{5}{4}, \frac{1}{2}, 2)$$
What have I done wrong? I'd really appreciate tips / explanations exactly where I am wrong. 
 A: Fact. Let $\{v_1,v_2,v_3,v_4\}$
be the vertices of a tetrahedron $T$. Then the center of mass of $T$ is
$$
\overline x=\frac{1}{4}\bigl(v_1+v_2+v_3+v_4\bigr)
$$
Details of this proof can be found here.
Applying the result to your tetrahedron gives
\begin{align*}
\overline x
&= \frac{1}{4}\bigl(v_1+v_2+v_3+v_4\bigr) \\
&= \frac{1}{4}\bigl(2-1+1+3,0+1+0+1,1+1+2+4\bigr) \\
&= \frac{1}{4}(5,2,8)
\end{align*}
which matches your desired result.
Essentially, you need to weight each vertex evenly and you seem to have taken a weighted sum.
A: suppose first we locate the origin of coordinates at the centroid. assume the position of the centroid is a linear function of the vertices. by symmetry such a function must be
$$
\tau(x_1+x_2+x_3+x_4)
$$
if we uniformly expand the tetrahedron  by a factor $\lambda$ by linearity the coordinates of the centroid are also multiplied by $\lambda$. however the centroid remains at the origin. hence in this system of coordinates we must have
$$
\sum_1^4 x_k=0 \tag{1}
$$
now shift the origin by $a$. in the new system of coordinates the centroid is $-a$, and the vertices are $x_k-a$
hence
$$
\tau\sum_1^4 (x_k-a) =-a \tag{2}
$$
(1) and (2) imply $\tau=\frac14$
