# determine the points in yz -plane and xz-plane which are equidistant from points A(1,-1,0), B(2,1,2) and C (3, 2 ,-1).

let the point P(0,b,c) is in y-z plane such that its distances from the points A(1,-1,0) , B(2,1,2) and C(3,2,-1) are equal.
therefore
PA = PB and PB = PC

solving eq(1) and eq(2), we have:

therefore the required point in y-z plane is $\left(0,\frac{31}{16},-\frac{3}{16}\right)$
similarly by assuming the point Q(x,0,z) in x-z plane we can find the another point in x-z plane

hope this helps you

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