Given: Equation of sphere is x2+y2+z2+2x−2y−6z−14=0 By comparing the given equation with x2+y2+z2+2ux+2vy+2wz+d=0, we get ⇒u=1,v=−1,w=−3 and d=−14 As we know that, equation x2+y2+z2+2ux+2vy+2wz+d=0 represents a sphere with centre (−u,−v,−w) and radius r=√u2+v2+w2−d So, the centre of given sphere is: (- 1, 1, 3) By substituting u=1,v=−1,w=−3 and d=−14 in the equation r=√u2+v2+w2−d we get, ⇒r=√12+(−1)2+(−3)2−(−14)=5 Hence, the centre is (- 1, 1, 3) and radius is 5 units