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.
PA = PB and PB = PC
PA2=PB2 and PB2=PC212+(b+1)2+c2=22+(b-1)2+(c-2)21+2b+1=4-2b+1-4c+42b+2=-2b-4c+94b+4c=7 ..........(1)22+(b-1)2+(c-2)2=32+(b-2)2+(c+1)24-2b+1-4c+4=9-4b+4+2c+19-2b-4c=9-4b+2c+52b-6c=5 .............(2)
solving eq(1) and eq(2), we have:
4b+4c=74b-12c=1016c=-3c=-316and 2b=6c+52b=6*(-316)+52b=-98+5=318b=3116
therefore the required point in y-z plane is 0,3116,-316
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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