What I tried:
Let $A(2,0,4)$, $B(4,1,-1)$, $C(6,7,7)$ then
$$\vec{AB}=(2,1,-5), \vec{AC}=(4,7,3), \vec{BC}=(2,6,8)$$
Then I calculated the angle between vectors:
$$\begin{aligned} \alpha_1 &= \cos^{-1}\left(\frac{(2,1,-5)(4,7,3)}{\sqrt{2^2+1^2+(-5)^2}\sqrt{4^2+7^2+3^2}}\right) \\ &= \cos^{-1}(0)=90° \\ \alpha_2 &= \cos^{-1}\left(\frac{(4,7,3)(2,6,8)}{\sqrt{4^2+7^2+3^2}\sqrt{2^2+6^2+8^2}}\right) \\ &= \cos^{-1}\left(\frac{74}{\sqrt{74}\sqrt{104}}\right)=32.49\\ \alpha_3 &= \cos^{-1}\left(\frac{(2,6,8)(2,1,-5)}{\sqrt{2^2+6^2+8^2}\sqrt{2^2+1^2+(-5)^2}}\right) \\ &= \cos^{-1}\left(\frac{-30}{\sqrt{104}\sqrt{30}}\right)=122.5° \end {aligned}$$
As you can see, these angles don't even form a triangle, what am I doing wrong, any thoughts?