Problem source

The plane
\[
{x \over a} + {y \over b} +{z \over c} = 1
\]
meets the co-ordinate axes at the points $A$, $B$ and $C\,$. The point $M$ has coordinates $\left( \frac12 a, \frac12 b, \frac 12 c \right)$ and $O$ is the origin.
Show that $OM$ meets the plane at the centroid  $\left( \frac13 a, \frac13 b, \frac 13 c \right)$ of triangle $ABC$. Show also that the perpendiculars to the plane from $O$ and from $M$ meet the plane at the orthocentre and at the circumcentre of triangle $ABC$ respectively.
Hence prove that the centroid of a triangle lies on the line segment joining its orthocentre and circumcentre, and that it divides this line segment in the ratio $2 : 1\,$. 
[The \textit{orthocentre} of a triangle is the point at which the three altitudes intersect; the \textit{circumcentre} of a triangle is the point equidistant from the three vertices.]

Solution source

The line $OM$ is $\lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix}$, then we need $1 = \lambda \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} \frac1{a} \\ \frac1b \\ \frac1c \end{pmatrix} = 3 \lambda \Rightarrow \lambda = \frac13$. Therefore $OM$ meets the plane at the centroid.

The orthocentre is the point $\mathbf{h}$ such that 
$(\mathbf{a}-\mathbf{b}) \cdot (\mathbf{c} - \mathbf{h}) = 0 \Leftrightarrow  \begin{pmatrix} a \\ -b \\ 0 \end{pmatrix} \cdot  \begin{pmatrix} -p \\ -q \\ c-r \end{pmatrix} \Leftrightarrow ap-bq = 0$
$(\mathbf{b}-\mathbf{c}) \cdot (\mathbf{a} - \mathbf{h}) = 0 \Leftrightarrow  \begin{pmatrix} 0 \\ b \\ -c \end{pmatrix} \cdot  \begin{pmatrix} a-p \\ -q \\ -r \end{pmatrix} \Leftrightarrow bq-cr = 0$
$(\mathbf{c}-\mathbf{a}) \cdot (\mathbf{b} - \mathbf{h}) = 0 \Leftrightarrow  \begin{pmatrix} -a \\ 0 \\ c \end{pmatrix} \cdot  \begin{pmatrix} -p \\ b-q \\ -r \end{pmatrix} \Leftrightarrow cr-ap = 0$

ie $ap = bq = cr$ but this is clearly on the line $\lambda \begin{pmatrix} \frac1{a} \\ \frac1b \\ \frac1c \end{pmatrix}$ therefore the orthocentre is on the perpendicular from $O$

$M-A = \begin{pmatrix} -a/2 \\ b/2 \\ c/2 \end{pmatrix}$ so $|M-A|=|M-B|=|M-C|$

Also by pythagoras the point of intersection satisfies $|M-P|^2 + |P-A|^2 = |M-A|^2$ so $|P-A|^2 = |P-B|^2 = |P-C|^2$, therefore $P$ is the circumcentre.

Since all these points are in the same plane and $OGM$ is a line, we have the points are in a line. Similar triangles gives the desired ratio