Answer
$ \dfrac{x \sqrt 3}{2}-\dfrac{y}{2}+z=\dfrac{\pi}{3}$
Work Step by Step
We need a point on the plane to calculate the tangent plane and the normal vector to the plane. For this, we need to take the cross product of two vectors in the plane.
The normal vector to the tangent plane becomes : $r(u,v)=-\sin v i-\cos vj+u k$
Let us consider that $a$ is the position vector of a point on the plane and $n$ is the vector normal to the plane, then the equation of plane is given by: $(r-a) \cdot n=0$ and for the specified points $(1, \dfrac{\pi}{3})$, we have:
$r(1, \dfrac{\pi}{3})=\dfrac{1}{2} i+ \dfrac{\sqrt 3}{2} j+\dfrac{\pi}{3} k$
$\implies (x-\dfrac{1}{2}) \cdot (\dfrac{\sqrt 3}{2})+(y-\dfrac{\sqrt 3}{2}) \cdot (-\dfrac{-1}{2}) +(z-\dfrac{\pi}{3}) \cdot (1)=0$
$\implies \dfrac{x \sqrt 3}{2}-\dfrac{y}{2}+z=\dfrac{\pi}{3}$
