#### Answer

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

#### Work Step by Step

Here, the level curve for $f(x,y,z)=\ln (x^2+y+z^2)$ has the form of $c=\ln (x^2+y+z^2)$
Also, $x=-1;y=2;z=1$
Then $c=\ln [(-1)^2+(2)+(1)^2]$
or,$ c= \ln 4 $
Now, $c=\ln (x^2+y+z^2) \implies \ln 4=\ln (x^2+y+z^2)$
or, $e^{(\ln 4)}=e^{[\ln (x^2+y+z^2)]}$
Hence, $(x^2+y+z^2)=4$