Answer
a. $x+y+z=3$ b. $x=1+2t,y=1+2t; z=1+2t$
Work Step by Step
a. As we know that the vector equation is given by:$r(x,y,z)=r_0+t \nabla f(r_0)$
The equation of tangent line is given as: $\nabla f(1,1,1)=\lt 2,2,2 \gt$
Now, $2(x-1)+2(y-1)+2(z-1)=0$
or, $2x+2y+2z=6 $ This implies that $x+y+z=3$
b. As we know that the vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$
The parametric equations for $\nabla f(1,1,1)=\lt 2,2,2 \gt$ are as follows:
$x=1+2t,y=1+2t; z=1+2t$
