Answer
$\dfrac{x \sqrt 3}{2}-\dfrac{y}{2}+z=\dfrac{\pi}{3}$
Work Step by Step
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: $(r-a) \cdot n=0$
The normal vector tangent to the plane is : $n=-\sin v i-\cos vj+u k$ and the point $(1, \dfrac{\pi}{3})$ corresponds to the parameter values as: $n=\dfrac{1}{2} i+ \dfrac{\sqrt 3}{2} j+\dfrac{\pi}{3} k$
Now, $(r-a) \cdot n=0$
$\implies (x-\dfrac{1}{2}) \cdot (\dfrac{\sqrt 3}{2})+(y-\dfrac{\sqrt 3}{2}) \cdot (-\dfrac{-1}{2}) +(z-\dfrac{\pi}{3}) \cdot (1)=0$
$\implies \dfrac{x \sqrt 3}{2}-\dfrac{y}{2}+z=\dfrac{\pi}{3}$
