solution: We see that $(3,-10,1)$ is a point on the first line.
  
  A point on the second line is given by
  \[\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix} + t \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix} = \begin{pmatrix} -5 + 4t \\ -3 - 18t \\ 6 - 4t \end{pmatrix}.\][asy]
  unitsize (0.6 cm);
  
  pair A, B, C, D, E, F, H;
  
  A = (2,5);
  B = (0,0);
  C = (8,0);
  D = (A + reflect(B,C)*(A))/2;
  
  draw(A--D);
  draw((0,5)--(8,5));
  draw((0,0)--(8,0));
  
  dot("$(3,-10,1)$", A, N);
  dot("$(-5 + 4t, -3 - 18t, 6 - 4t)$", D, S);
  [/asy]
  
  The vector pointing from $(3,-10,1)$ to $(-5 + 4t, -3 - 18t, 6 - 4t)$ is then
  \[\mathbf{v} = \begin{pmatrix} -8 + 4t \\ 7 - 18t \\ 5 - 4t \end{pmatrix}.\]For the point on the second line that is closest to $(3,-10,1),$ this vector will be orthogonal to the direction vector of the second line, which is $\begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}.$  Thus,
  \[\begin{pmatrix} -8 + 4t \\ 7 - 18t \\ 5 - 4t \end{pmatrix} \cdot \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix} = 0.\]This gives us $(-8 + 4t)(4) + (7 - 18t)(-18) + (5 - 4t)(-4) = 0.$  Solving, we find $t = \frac{1}{2}.$
  
  Substituting this value into $\mathbf{v},$ we find that the distance between the parallel lines is then
  \[\|\mathbf{v}\| = \left\| \begin{pmatrix} -6 \\ -2 \\ 3 \end{pmatrix} \right\| = \boxed{7}.\]
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  One line is defined by
  \[\begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}.\]Another line is defined by
  \[\begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix} + u \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}.\]These two lines are parallel.  Find the distance between these two lines.
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 7
extra_info.question: One line is defined by
  \[\begin{pmatrix} 3 \\ -10 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -9 \\ -2 \end{pmatrix}.\]Another line is defined by
  \[\begin{pmatrix} -5 \\ -3 \\ 6 \end{pmatrix} + u \begin{pmatrix} 4 \\ -18 \\ -4 \end{pmatrix}.\]These two lines are parallel.  Find the distance between these two lines.
extra_info.split: test