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