Answer
$x=1; y=1+2t; z=1-2t$
Work Step by Step
The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$
Given: $f(x,y,z)=x+y^2+z-4$
The equation of tangent line $v=0 i +2j-2k$ is given as follows:
Now, we have the parametric equations for $\nabla f(1,1,1)=\lt 1,2,2 \gt$ are:
$x=1; y=1+2t; z=1-2t$
