solution: In general,
  \[\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\]is the signed area of the triangle with vertices at $(x_1,y_1),$ $(x_2,y_2),$ and $(x_3,y_3).$  (The area is signed, i.e. it can be positive or negative, depending on the orientation of the triangle.)  Here, the sides of the triangle are 3, 4, and 5, which is a right triangle.  Therefore, its area is $\frac{1}{2} \cdot 3 \cdot 4 = 6.$  Then
  \[\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm 12,\]so
  \[\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}^2 = \boxed{144}.\]
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  Let $x_1,$ $x_2,$ $x_3,$ $y_1,$ $y_2,$ and $y_3$ be real numbers such that
  \begin{align*}
  (x_1 - x_2)^2 + (y_1 - y_2)^2 &= 9, \\
  (x_1 - x_3)^2 + (y_1 - y_3)^2 &= 16, \\
  (x_2 - x_3)^2 + (y_2 - y_3)^2 &= 25.
  \end{align*}Find $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}^2.$
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 144
extra_info.question: Let $x_1,$ $x_2,$ $x_3,$ $y_1,$ $y_2,$ and $y_3$ be real numbers such that
  \begin{align*}
  (x_1 - x_2)^2 + (y_1 - y_2)^2 &= 9, \\
  (x_1 - x_3)^2 + (y_1 - y_3)^2 &= 16, \\
  (x_2 - x_3)^2 + (y_2 - y_3)^2 &= 25.
  \end{align*}Find $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}^2.$
extra_info.split: test