Here's an elegant solution. Near the point {2,3,2}, x^2 can be approximated by 2x, y^2 by 3y and z^2 by 2z. Therefore
4x^2+3y^2+z^2=47 near the point reduces to 8x+9y+2z=47, which is the equation of the tangent plane. To find the normal vector to the plane at {2,3,2}, rewrite the equation for the plane to the form a(x-2)+b{y-3)+c(z-2)=0, The normal vector is then {a,b,c}.