Given the vectors (1, 2, 3) and (-1, 1, 1) and (2,3, - 6) , find the vector that makes the same angle with ...?

Let u,v,w be unit vectors in the given directions.

If d is the required direction then we need d•u = d•v = d•w

These equations imply d•(u−v) = 0 and d•(u−w) = 0

Hence d is perp to both u−v and u−w and so in direction (u−v)X(u−w)

Expanding cross product gives d = uXu – uXw – vXu + vXw = uXv + vXw + wXu

You can see it works because d•u = u•(vXw), d•v = v•(wXu), d•w = w•(uXv) and there is equality since the three triple scalar products are equal. (tetrahedron volume)

For the case in point u=(1,2,3)/√14, v=(−1,1,1)/√3, w=(2,3,−6)/7

d = (u−v)X(u−w) = ( (1,2,3)/√14 − (−1,1,1)/√3 ) X ( (1,2,3)/√14 − (2,3,−6)/7 )

Wolfram gives d ≈ ( −0.0948271, −1.40529, 0.0886972 ), |d| ≈ 1.411275758306

To find angle compute d•v = −(16√42)/147 = |d|cosθ = 1.411275758306cosθ

∴ cosθ = −0.49982205 → acute angle = 60.0117°

Just to check angle I redid with exact d = (1/7)√{ (2/21)(1185+105√3+35√14–73√42) }

Which gives cosθ = −16/√{1185+105√3+35√14−73√(42)} =0.499822050036 so θ≠60