[tex]\vec F(x,y,z)=\dfrac74\cos(xyz)\langle yz,xz,xy\rangle[/tex]
Computing the line integral directly is cumbersome, if not impossible by elementary means. Let's instead try to determine if [tex]\vec F[/tex] is conservative. We look for a scalar function [tex]f(x,y,z)[/tex] such that [tex]\nabla f=\vec F[/tex]. We should have
[tex]\dfrac{\partial f}{\partial x}=yz\cos(xyz)[/tex]
(ignoring the 7/4 for a moment). Integrating both sides wrt [tex]x[/tex] gives
[tex]\displaystyle\int\cos(xyz)yz\,\mathrm dx=\sin(xyz)+g(y,z)[/tex]
Then differentiating wrt [tex]y[/tex] gives
[tex]\dfrac{\partial(\sin(xyz))}{\partial y}=xz\cos(xyz)=xz\cos(xyz)+\dfrac{\partial g}{\partial y}[/tex]
[tex]\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)[/tex]
Differentiating wrt [tex]z[/tex] gives
[tex]\dfrac{\partial(\sin(xyz))}{\partial z}=xy\cos(xyz)=xy\cos(xyz)+\dfrac{\mathrm dh}{\mathrm dz}[/tex]
[tex]\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C[/tex]
So we have (and here we re-introduce the 7/4)
[tex]f(x,y,z)=\dfrac74\sin(xyz)+C[/tex]
and by the fundamental theorem of calculus,
[tex]\displaystyle\int_C\nabla f\cdot\mathrm d\vec r=f(\vec b)-f(\vec a)[/tex]
where [tex]\vec a[/tex] and [tex]\vec b[/tex] are vectors representing the start- and endpoints of [tex]C[/tex]. So
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\frac74\sin\frac\pi6-\frac74\sin\frac\pi2=\boxed{\frac78}[/tex]
