Triangle defined by one point

Let $P$ denote the point of origin, $A,B,C$ vertices of the triangle, $R_{c}$ its circumradius and $a,b,c$ its side lengths.

$\text{Tetrahedron }ABCP \text{ is trirectangular } \Rightarrow \text{ orthogonal }$ $\Rightarrow PH \perp \triangle ABC$ .

Plane on which $\triangle ABC$ lies can be found $x + 2y + 3z = 14$ and points $A(14,0,0), B(0,7,0), C(0,0,14/3)$ .

Triangle's side lengths can be found using Pythagorean theorem, and the rest of the values using regular formulas. Heron's formula or de Gua's theorem to find triangle area and then:

$R_{c}=\frac{abc}{4 \triangle ABC}$

$OH^2 = 9 R_{c}^2 - (a^2 + b^2 + c^2)$

$R = R_{c}/2 , ON = OH/2$

Note:

As an additional exercise I got the following:

For $H(k,l,m)$ plane has equation $kx + ly + mz = k^2 + l^2 + m^2$ .

Vertices: $A(a_{l},0,0) B(0,b_{l},0) C(0,0,c_{l})$ $a_{l}=\frac{k^2 + l^2 + m^2}{k}, b_{l}=\frac{ k^2 + l^2 + m^2}{l}, c_{l}=\frac{k^2 + l^2 + m^2}{m})$

Centroid: $G(\frac{k^2 + l^2 + m^2}{3k}, \frac{k^2 + l^2 + m^2}{3l}, \frac{ k^2 + l^2 + m^2}{3m})$

Circumcentre : $O ( \frac{l^2+m^2}{2k} , \frac{k^2+m^2}{2l} , \frac{k^2+l^2}{2m} ) = O( \frac{a_{l}-k}{2}, \frac{b_{l}-l}{2}, \frac{c_{l}-m}{2})$

Circumradius: $R_{c}^2=((a_{l}^2 + b_{l}^2 + c_{l}^2) - (k^2+l^2+m^2))/4$

$OH^2=(a_{l}^2 + b_{l}^2 + c_{l}^2 - 9(k^2 + l^2+ m^2))/4$

$R_{c}^2 - OH^2 = 2(k^2 + l^2+ m^2) = 2 HP^2$

$R_{c}^2 + OH^2=2 OP^2$

$a_{l} k = b_{l} l = c_{l} m = k^2 + l^2+ m^2$