Answer
$$-3$$
Work Step by Step
The line equation can be written as:
$ r(t)=r_0+kt=(1,1,1)+t \space \lt 1,2,-2 \gt=\lt 1+t, 1+2t, 1-2t \gt $
or, $x=1+t$
or, $ dx= dt \\ y=1+2t $
or, $ dy= 2 dt \\ z= 1-2t$
or, $ dz= -2dt $
Now, we will substitute all the above values in the given integral.
$ \int_{0,1,1}^{2,3,-1} \ y dx +x \ dy +4 \ dz=\int_0^1 (1+2t) dt +(1+t) 2dt -8 dt=\int_0^1 (4t-5) dt$
or, $= [2t^2 -5t])_0^1 $
or, $=-3$