[tex]\displaystyle\int_{\mathcal C}2xyz\,\mathrm dx+x^2z\,\mathrm dy+x^2y\,\mathrm dz=\int_{\mathcal C}\underbrace{(2xyz,x^2z,x^2y)}_{\mathbf F}\cdot\underbrace{(\mathrm dx,\mathrm dy,\mathrm dz)}_{\mathrm d\mathbf r}[/tex]
The fact that [tex]\mathcal C[/tex] is not described exactly is a clue to try to show the vector field [tex]\mathbf F[/tex] is conservative. That means you need to show there is some scalar function [tex]f[/tex] for which [tex]\mathbf F=\nabla f[/tex]; that is, the following set of PDEs is satisfied:
[tex]\dfrac{\partial f}{\partial x}=2xyz[/tex]
[tex]\dfrac{\partial f}{\partial y}=x^2z[/tex]
[tex]\dfrac{\partial f}{\partial z}=x^2y[/tex]
Integrating both sides of the first PDE with respect to [tex]x[/tex] gives
[tex]f(x,y,z)=x^2yz+g(y,z)[/tex]
Differentiating with respect to [tex]y[/tex] gives
[tex]\dfrac{\partial f}{\partial y}=x^2z+\dfrac{\partial g}{\partial z}=x^2z\implies\dfrac{\partial g}{\partial z}=0\implies g(y,z)=h(z)[/tex]
Differentiating with respect to [tex]z[/tex] gives
[tex]\dfrac{\partial f}{\partial z}=x^2y+\dfrac{\mathrm dh}{\mathrm dz}=x^2y\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C[/tex]
So the potential scalar function is
[tex]f(x,y,z)=x^2yz+C[/tex]
By the fundamental theorem of calculus, we have for any oriented simple curve [tex]\mathcal C[/tex] parameterized by a vector-valued function [tex]\mathbf r[/tex], starting at [tex]\mathbf a[/tex] and terminating at [tex]\mathbf b[/tex],
[tex]\displaystyle\int_{\mathcal C}\nabla f\cdot\mathrm d\mathbf r=f(\mathbf r(\mathbf b))-f(\mathbf r(\mathbf a))[/tex]
In this case, [tex]\mathbf a=(1,2,4)[/tex] and [tex]\mathbf b=(1,4,16)[/tex], so the line integral has a value of
[tex]\displaystyle\int_{\mathcal C}2xyz\,\mathrm dx+x^2z\,\mathrm dy+x^2y\,\mathrm dz=f(1,4,16)-f(1,2,4)=64-8=56[/tex]
