Answer
$ 3x+4y-12z+13=0$
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 : $n=-3v^2 i-2uj+6uv^2 k$ and the points (5, 2,3) corresponds to the parameter values is given by: $n=-3 i-4j+12 k$
$(r-a) \cdot n=0$
This implies that $(x-5) \cdot (-3)+(y-2) \cdot (-4) +(z-3) \cdot (12)=0$
and $ -3x-4y+12z-13=0$
So, $ 3x+4y-12z+13=0$
