Answer
$$3x+4y-12z+13=0$$
Work Step by Step
The normal vector tangent to the plane is:
$n=-3v^2 i-2uj+6uv^2 k$ .
The point (5, 2,3) corresponds to the parameter values: $n=-3 i-4j+12 k$
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 can be written as: $(r-a) \cdot n=0$
$$(x-5) \cdot (-3)+(y-2) \cdot (-4) +(z-3) \cdot (12)=0 \\-3x-4y+12z-13=0\\ 3x+4y-12z+13=0$$
