solution: From the distance formula, we compute that $AB = 3 \sqrt{6},$ $AC = 9 \sqrt{2},$ and $BC = 3 \sqrt{6}.$  Then from the Law of Cosines,
  \[\cos \angle ABC = \frac{(3 \sqrt{6})^2 + (3 \sqrt{6})^2 - (9 \sqrt{2})^2}{2 \cdot 3 \sqrt{6} \cdot 3 \sqrt{6}} = -\frac{1}{2}.\]Therefore, $\angle ABC = \boxed{120^\circ}.$
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  Let $A = (1,-11,2),$ $B = (3,-4,1),$ and $C = (-2,1,-1).$  Compute $\angle ABC,$ in degrees.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 120^\circ
extra_info.question: Let $A = (1,-11,2),$ $B = (3,-4,1),$ and $C = (-2,1,-1).$  Compute $\angle ABC,$ in degrees.
extra_info.split: test