Answer
$2 \sqrt 3 x+12y-6 \sqrt 3 z=\sqrt 3$
Work Step by Step
The normal vector tangent to the plane is:
$n=-\sin u \sin v \cos vi-\cos u \cos vj+\cos^2 u \cos v k$
and the point $( \dfrac{\pi}{6}, \dfrac{\pi}{6})$ corresponds to the parameter values as: $n=\dfrac{\sqrt 3}{8} i+ \dfrac{ 3}{4} j+\dfrac{ 3 \sqrt 3}{8} k$
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, $(x-\dfrac{1}{2}) \cdot (\dfrac{-\sqrt 3}{8})+(y-\dfrac{\sqrt 3}{4}) \cdot (-\dfrac{3}{4}) +(z-\dfrac{1}{2}) \cdot (3\sqrt 3/8)=0$
$\implies 2 \sqrt 3 x+12y-6 \sqrt 3 z=\sqrt 3$
