NDA Math Three Dimensional Geometry

Let equation of a sphere be
(x−a)2+(y−b)2+(z−c)2=r2
If the origin and given points lie on the sphere, then
a2+b2+c2=r2......(1)
From 1 and 2
2a+b−c=3...(5)
From 1 and 3
a+5b−4c=21....(6)
From 1 and 4
−a+2b−3c=14....(7)
On solving, the point (−1,2,−3) satisfies the above equations
⇒r=√14
The equation of the sphere is
(x−1)2+(y−2)2+(z+3)2=14
∴(0,4,0) lies on the sphere
Distance between (1,1,1) and centers (−1,2,−3)
Hence, statement 1 is correct and statement 2 is incorrect.