 Orthogonal projection onto an intercept plane

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 $\mathbf{d}=\mathbf{n}/\Vert \mathbf{n}{\Vert }^2$, with distance $1/\Vert \mathbf{n}\Vert$. 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 $1/\Vert \mathbf{n}{\Vert }^2$. 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.