Consider the points 
in 
. Determine the cosines of the angles of the triangle whose vertices are on these points.
Let 
denote the angle between 
and 
, let 
denote the angle between 
and and let 
denote the angle between and . Then we compute the cosines,
![\begin{align*} \cos a &= \frac{(2,-1,1) \cdot (1,-3,-5)}{\sqrt{6} \sqrt{35}} = 0 \\[9pt] \cos b &= \frac{(1,-3,-5)\cdot(3,-4,-4)}{\sqrt{35} \sqrt{41}} = \frac{35}{\sqrt{35}\sqrt{41}} = \sqrt{\frac{35}{41}} \\[9pt] \cos c &= \frac{(2,-1,1) \cdot (3,-4,-4)}{\sqrt{6} \sqrt{41}} = \frac{6}{\sqrt{6} \sqrt{41}} = \sqrt{\frac{6}{41}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f47d296462e8647f2842758d3a3a72ae_l3.png "Rendered by QuickLaTeX.com")
The answer is incorrect here. As Refael says, one needs to take differences between vectors. Then you also need to align vectors in such a way that every pair of vectors originates from the same point: BC vs BA, AB vs AC, and CA vs CB: the sum of the angles has to be pi in a triangle.
That reasoning is wrong. Let A, B and C the given vectors. So, what we want is the angle between the vectors B-C and A-C, B-A and C-A, C-B and A-B.