= Orthogonal projection onto an intercept plane
{title2=$\mathbf d=\mathbf n/(\mathbf n\cdot\mathbf n),\quad\mathbf n=(1/a,1/b,1/c)$}
For positive coordinate intercepts $A=(a,0,0)$, $B=(0,b,0)$ and $C=(0,0,c)$, their <plane equation> is $\mathbf n\cdot\mathbf r=1$. The <closest point on a plane> to the origin is $\mathbf d=\mathbf n/\|\mathbf n\|^2$, with distance $1/\|\mathbf n\|$. Its <barycentric coordinate system> coefficients are proportional to $a^{-2},b^{-2},c^{-2}$ and sum to one.
The projection is also the <orthocenter> of triangle $ABC$. Indeed $\mathbf a\cdot(\mathbf b-\mathbf c)=0$, and $\mathbf d\cdot\mathbf b=\mathbf d\cdot\mathbf c$ because each equals $1/\|\mathbf n\|^2$. Hence $(\mathbf a-\mathbf d)\cdot(\mathbf b-\mathbf c)=0$. Cycling the vertices gives all three altitude directions. This connects an ambient-space <orthogonal projection> with intrinsic triangle geometry.
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