Find the point in the $xz$-plane that is equidistant from the points $(1,-1,0),$ $(2,1,2),$ and $(3,2,-1).$
Question:
Find the point in the $xz$-plane that is equidistant from the points $(1,-1,0),$ $(2,1,2),$ and $(3,2,-1).$
Answer:
Let's think step by step.

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\left( \frac{31}{10}, 0, \frac{1}{5} \right)
Find the point in the $xz$-plane that is equidistant from the points $(1,-1,0),$ $(2,1,2),$ and $(3,2,-1).$
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