(grad)Ø=∂Ø/∂xi+∂Ø/dyj+∂Ø/∂zk, so Ø=x^2yz^3+C where C is a constant, because ∂Ø/∂x=2xyz^3, ∂Ø/∂y=x^2z^3 and ∂Ø/dz=3x^2yz^2, then Ø(1,-2,2)=4=1*(-2)2^3+C=-16+C, and C=20. Ø=x^2yz^3+20.
[Integrating ∂Ø/∂y we get Ø=x^2z^3y+C which also fits other partial differentials.]