[tex]\vec f(x,y,z)=yze^{xz}\,\vec\imath+e^{xz}\,\vec\jmath+xye^{xz}\,\vec k[/tex]
We're looking for a scalar function [tex]f[/tex] such that [tex]\nabla f=\vec f[/tex]. This means
[tex]\dfrac{\partial f}{\partial x}=yze^{xz}[/tex]
[tex]\dfrac{\partial f}{\partial y}=e^{xz}[/tex]
[tex]\dfrac{\partial f}{\partial z}=xye^{xz}[/tex]
In the second PDE, integrate both side with respect to [tex]y[/tex]:
[tex]\dfrac{\partial f}{\partial y}=e^{xz}\implies f(x,y,z)=ye^{xz}+g(x,z)[/tex]
Differentiate both sides of this with respect to [tex]x[/tex]:
[tex]yze^{xz}=yze^{xz}+\dfrac{\partial g}{\partial x}[/tex]
[tex]\dfrac{\partial g}{\partial x}=0\implies g(x,z)=h(z)[/tex]
[tex]\implies f(x,y,z)=ye^{xz}+h(z)[/tex]
Differentiate this with respect to [tex]z[/tex]:
[tex]xye^{xz}=xye^{xz}+\dfrac{\mathrm dh}{\mathrm dz}[/tex]
[tex]\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=K[/tex]
for some constant [tex]K[/tex], so that
[tex]f(x,y,z)=ye^{xz}+K[/tex]
By the fundamental theorem of calculus, this line integral has a value of
[tex]\displaystyle\int_C\nabla f\cdot\mathrm d\vec r=f(\vec r(5))-f(\vec r(0))=f(30,22,20)-f(5,-3,0)=\boxed{22e^{600}+3}[/tex]
since
[tex]\vec r(5)=30\,\vec\imath+22\,\vec\jmath+20\,\vec k[/tex]
[tex]\vec r(0)=5\,\vec\imath-3\,\vec\jmath[/tex]
