Orthogonal tetrahedron

Let $E,F$ denote feet of the altitudes from $B,C$ respectively.

Tetrahedron's net looks like this:

The height of the tetrahedron is the geometric mean of $AH$ and $HD$ . $h=\sqrt{AH \times HD}=6$ .

$AH \times HD = BH \times HE \Rightarrow HE = 6$ .

$\dfrac{HD}{AD} \dfrac{HE}{BE}\dfrac{HF}{CF}=1 \Rightarrow HF=6 \dfrac{\sqrt{21}}{7}, CH = 2 \sqrt{21} , CF = 20 \dfrac{\sqrt{21}}{7}$

$h_{a}=15, h_{b}=12, h_{c}=20 \dfrac{\sqrt{21}}{7}$

To calculate area of $\triangle ABC$ we can use the formula using altitudes:

$H_{r} = ( h_{a} ^{-1} + h_{b} ^{-1} + h_{c} ^{-1})/2$

$Area ^{-1} = 4 \sqrt{H_{r} (H_{r}-h_{a} ^{-1})(H_{r}-h_{b} ^{-1})(H_{r}-h_{c} ^{-1}}) \Rightarrow \triangle ABC = 3 \sqrt{60}$

$V = \dfrac{ 6 \times 3 \sqrt{60}}{3} = 120 \sqrt{3} \approx 207.85$