Show that $(2,0,4) , (4,1,-1) , (6,7,7)$ form a right triangle

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?

It's enough to show the following: $$(2,1,-5)\cdot(4,7,3)=0$$ and we are done!

The distances satisfy the Pythagorean theorem.

$d(A, B) = \sqrt{2^2 + 1^2 + 5^2} = \sqrt{30}$

$d(A, C) = \sqrt{4^2 + 7^2 + 3^2} = \sqrt{74}$

$d(B, C) = \sqrt{2^2 + 6^2 + 8^2} = \sqrt{104}$

And indeed:

$\sqrt{30}^2 + \sqrt{74}^2 = \sqrt{104}^2$

Therefore it is a right triangle (link).