Answer
$x-2z+1=0$
Work Step by Step
The normal vector tangent to the plane is:
$n=-2 u i+2u^2 \sin v j+4 u^2 \cos v k$
The point $(1,0)$ corresponds to the parameter values as: $n=-2i+0 j+4k$
and $r(1,0) =i+0 j+k$
When $a$ is the position vector of a point on the plane and $n$ is the vector normal to the plane, then the equation of the plane is: $(r-a) \cdot n=0$
Therefore, $(x-1) \cdot (-2)+(y-0) \cdot (0) +(z-1) \cdot (4)=0$
$\implies x-2z+1=0$
