Computing the line integral directly would be easy enough, but in general it's worth checking to see if the vector field is conservative; that is, whether there exists a scalar function [tex]f[/tex] whose gradient corresponds exactly to the given vector field [tex]\mathbf f[/tex], or [tex]\nabla f=\mathbf f[/tex]. Equivalently, we're looking for [tex]f[/tex] such that
[tex]\dfrac{\partial f}{\partial x}=yz[/tex]
[tex]\dfrac{\partial f}{\partial y}=xz[/tex]
[tex]\dfrac{\partial f}{\partial z}=xy[/tex]
We find that
[tex]f_x=yz\implies f(x,y,z)=xyz+g(y,z)[/tex]
[tex]f_y=xz+g_y=xz\implies g_y=0\implies g(y,z)=h(z)[/tex]
[tex]f_z=xy+h_z=xy\implies h_z=0\implies h(z)=C[/tex]
and so there is indeed such a function [tex]f[/tex], with [tex]f(x,y,z)=xyz+C[/tex].
Thus by the fundamental theorem of calculus, the line integral is path-independent, and
[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{\mathcal C}\nabla f\cdot\mathrm d\mathbf r=f(\mathbf b)-f(\mathbf a)[/tex]
where [tex]\mathcal C[/tex] is any path starting at [tex]\mathbf a[/tex] and ending at [tex]\mathbf b[/tex]. Here,
[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(9,7,4)-f(0,0,0)=252[/tex]
