#### Answer

$(x^2+y+z^2)=4$

#### Work Step by Step

Remember that the level curve for $f(x,y,z)=\ln (x^2+y+z^2)$ has the form of $c=\ln (x^2+y+z^2)$ ...(1)
Here, we have $x=-1,y=2,z=1$
Then $c=\ln ((-1)^2+(2)+(1)^2) \implies c= \ln 4 $
Thus, equation (1), becomes: $\ln 4=\ln (x^2+y+z^2)$
This implies that $e^{\ln 4}=e^{ \ln (x^2+y+z^2)}$
Hence, $(x^2+y+z^2)=4$