Answer
$x-y+2z=1$
Work Step by Step
As we are given that $(y-x)^{1/2}-z=0$
Since, we have the vector equation $r(x,y,z)=r_0+t \nabla f(r_0)$
The equation of the tangent line for $\nabla f(1,2,1)=\lt \dfrac{-1}{2},\dfrac{1}{2},1 \gt$ is
$\dfrac{-1}{2}(x-1)+\dfrac{1}{2}(y-2)-(z-1)=0$
or, $\dfrac{-1}{2}x+\dfrac{1}{2}y-z=\dfrac{1}{2}$
or, $x-y+2z=1$
