Let O(x,y,z) be the circumcentre of ΔABC. OA=OB=OC Thus, OA=OB OA2=OB2 (x−1)2+(y−2)2+(z−3)2=(x−3)2+(y+1)2+(z−5)2 4x−6y+4z−21=0 Similarly, OB2=OC2 (x−3)2+(y+1)2+(z−5)2=(x−4)2+(y−0)2+(z+3)2 x+y−8z+5=0 Similarly OA2=OC2 (x−1)2+(y−2)2+(z−3)2=(x−4)2+(y−0)2+(z+3)2 6x−4y−12z−11=0 On solving the obtained equations, x=