Chapter 13 - Section 13.6 - Tangent Planes and Differentials - Exercises - Page 728: 15

$x=1-2t; y=1; z=\dfrac{1}{2}+2t$

As we are given that $f(x,y,z)=x^2+2y+2z-4=0$ Since, we have the vector equation $r(x,y,z)=r_0+t \nabla f(r_0)$ The equation of the tangent line is: $v=-2 i +2k$ Thus, the parametric equations for $\nabla f(1,1,\dfrac{1}{2})=\lt 2,2,2 \gt$ is $x=1-2t; y=1+0t=1; z=\dfrac{1}{2}+2t$ or, $x=1-2t; y=1; z=\dfrac{1}{2}+2t$