The correct option is B x2+y2+z2+2x+4y+6z−11=0
Equation to the sphere passes through the circle x2+y2+z2=5,x+2y+3z=3 is given by,
x2+y2+z2−5+λ(x+2y+3z−3)=0
⇒x2+y2+z2+λx+2λy+3λz−(3λ+5)=0
Given plane 4x+3y−15=0 touches this sphere,
∣∣
∣
∣∣4(−λ2)+3(−λ)−155∣∣
∣
∣∣=√(λ2)2+λ2+(3λ2+3λ+5)2
⇒5λ2−6λ−8=0⇒λ=2,−4/5
Hence, the sphere is x2+y2+z2+2x+4y+6z−11=0
There will be one more sphere corresponding to another value of λ.