solution: Let $P = (a,b,c).$  Then the distance from $P$ to the plane $x - z = 0$ is
  \[d_1 = \frac{|a - c|}{\sqrt{1^2 + (-1)^2}} = \frac{|a - c|}{\sqrt{2}}.\]The distance from $P$ to the plane $x - 2y + z = 0$ is
  \[d_2 = \frac{|a - 2b + c|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|a - 2b + c|}{\sqrt{6}}.\]And, the distance from $P$ to the plane $x + y + z = 0$ is
  \[d_3 = \frac{|a + b + c|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|a + b + c|}{\sqrt{3}}.\]Then the equation $d_1^2 + d_2^2 + d_3^2 = 36$ becomes
  \[\frac{(a - c)^2}{2} + \frac{(a - 2b + c)^2}{6} + \frac{(a + b + c)^2}{3} = 36.\]This simplifies to $a^2 + b^2 + c^2 = 36.$  Thus, $S$ is a sphere with radius 6, so its volume is
  \[\frac{4}{3} \pi \cdot 6^3 = \boxed{288 \pi}.\]
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  For a point $P,$ let $d_1,$ $d_2$ and $d_3$ represents the distances from $P$ to the planes $x - z = 0,$ $x - 2y + z = 0,$ and $x + y + z = 0.$  Let $S$ be the set of points $P$ such that
  \[d_1^2 + d_2^2 + d_3^2 = 36.\]Find the region of the volume enclosed by $S.$
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: 288 \pi
extra_info.question: For a point $P,$ let $d_1,$ $d_2$ and $d_3$ represents the distances from $P$ to the planes $x - z = 0,$ $x - 2y + z = 0,$ and $x + y + z = 0.$  Let $S$ be the set of points $P$ such that
  \[d_1^2 + d_2^2 + d_3^2 = 36.\]Find the region of the volume enclosed by $S.$
extra_info.split: test