Answer
$x=\dfrac{1}{2}-t; y=1; z=\dfrac{1}{2}+t$
Work Step by Step
The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$
Given: $f(x,y,z)=x+y^2+z-2=0$
The equation of tangent line is $v=- i +k$ or, $v=\lt -1,0,1 \gt$
Now, we have the parametric equations for $\nabla f(\dfrac{1}{2},1,\dfrac{1}{2})=\lt -1,0,1 \gt$ as follows:
$x=\dfrac{1}{2}-t; y=1+0t=1; z=\dfrac{1}{2}+t$
Thus, $x=\dfrac{1}{2}-t; y=1; z=\dfrac{1}{2}+t$
