[tex]\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k[/tex]
is conservative if there is a scalar function [tex]f(x,y,z)[/tex] such that [tex]\nabla f=\vec F[/tex]. This would require
[tex]\dfrac{\partial f}{\partial x}=y[/tex]
[tex]\dfrac{\partial f}{\partial y}=x[/tex]
[tex]\dfrac{\partial f}{\partial z}=3[/tex]
(or perhaps the last partial derivative should be 4 to match up with the integral?)
From these equations we find
[tex]f(x,y,z)=xy+g(y,z)[/tex]
[tex]\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)[/tex]
[tex]f(x,y,z)=xy+h(z)[/tex]
[tex]\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C[/tex]
[tex]f(x,y,z)=xy+3z+C[/tex]
so [tex]\vec F[/tex] is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints:
[tex]\displaystyle\int_{(1,2,3)}^{(5,7,-2)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz=\int_{(1,2,3)}^{(5,7,-2)}\nabla f(x,y,z)\cdot\mathrm d\vec r[/tex]
[tex]=f(5,7,-2)-f(1,2,3)=\boxed{18}[/tex]
