The answer is: #\frac{56\sqrt{77}}{3}\approx 163.8#

First, parameterize the line segment. The quickest way to do this is to let #P=(0,6,-1)# and #Q=(4,1,5)# and, thinking of these as vectors, find a vector going from #P# to #Q# by subtracting the components/coordinates of #P# from the corresponding components/coordinates of #Q# to get #\vec{v}=(4,-5,6)#. Now let #\vec(c)(t)=P+t\vec{v}=(4t,6-5t,-1+6t)# for #0\leq t\leq 1#.

The velocity vector is #\vec{c}'(t)=\vec{v}#, and its length (the speed) is #||\vec{v}||=\sqrt{16+25+36}=\sqrt{77}#. Letting #f(x,y,z)=x^2z#, the function we have to integrate as #t# goes from #t=0# to #t=1# is #f(4t,6-5t,-1+6t)||\vec{v}||=\sqrt(77)(4t)^2\cdot (-1+6t)=\sqrt{77}(96t^3-16t^2).#

(Note that #ds=\frac{ds}{dt}dt=||\vec{v}||dt#.)

Here's the integral calculation:

#\int_{0}^{1}\sqrt{77}(96t^3-16t^2)dt=\sqrt(77)(24t^{4}-\frac{16}{3}t^{3})\|_{t=0}^{t=1}#

#=\sqrt{77}\cdot \frac{72-16}{3}=\frac{56\sqrt{77}}{3}\approx 163.8#