Answer
$3x-y+3z=3$
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 plane is: $(r-a) \cdot n=0$
The normal vector tangent to the plane is : $n=-6u i+2j-6u k$ and the point (2,3,0) corresponds to the parameter values as: $n=-6 i+2j-6 k$
Now, $(r-a) \cdot n=0$
$\implies (x-2) \cdot (-6)+(y-3) \cdot (2) +(z-0) \cdot (-6)=0$
$\implies -6x+12+2y-6-6z=0$
$\implies 3x-y+3z=3$
