Answer
$$\dfrac{x \sqrt 3}{2}-\dfrac{y}{2}+z=\dfrac{\pi}{3}$$
Work Step by Step
In order to find the tangent plane, we need a point on the plane, and the normal vector to the plane. For this, we will take the cross product of two vectors in the plane.
The normal vector to the tangent plane is: $$r(u,v)=-\sin v i-\cos vj+u k$$
Now, for the specified points $(1, \dfrac{\pi}{3})$, we have:
$$r(1, \dfrac{\pi}{3})=\dfrac{i}{2}+ \dfrac{\sqrt 3}{2} j+\dfrac{\pi}{3} k$$
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: $(r-a) \cdot n=0$
$$ (x-\dfrac{1}{2}) \cdot (\dfrac{\sqrt 3}{2})+(y-\dfrac{\sqrt 3}{2}) \cdot (-\dfrac{-1}{2}) +(z-\dfrac{\pi}{3}) \cdot (1)=0 \\ \dfrac{x \sqrt 3}{2}-\dfrac{y}{2}+z=\dfrac{\pi}{3}$$
