Answer
$x=1+2t; y=1; z=1+2t$
Work Step by Step
As we are given that $f(x,y,z)=xyz; x^2+2y^2+3z^2=6$
Since, we have the vector equation $r(x,y,z)=r_0+t \nabla f(r_0)$
The equation of the tangent line is: $v=2 i -4j+2k=\lt 2,-4,2 \gt$
Thus, the parametric equations for $\nabla f(1,1,1)=\lt 1,1,1 \gt$ are
$x=1+2t; y=1+0t=1; z=1+2t$
or, $x=1+2t; y=1; z=1+2t$
