Answer
$3x-y+3z=3$
Work Step by Step
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: $(r-a) \cdot n=0$
The normal vector tangent to the plane becomes : $n=-6u i+2j-6u k$ and the points $(2,3,0)$ corresponds to the parameter values are: given by: $n=-6 i+2j-6 k$
$$(r-a) \cdot n=0 \\ (x-2) \cdot (-6)+(y-3) \cdot (2) +(z-0) \cdot (-6)=0 \\ -6x+12+2y-6-6z=0 \\ 3x-y+3z=3$$
