Answer
$$-2$$
Work Step by Step
We know that the line equation can be defined as: $$ r(t)=r_0+kt=(1,1,1)+t \space \lt 2, 4, -2 \gt \\=\lt 1+2t, 1+4t, 2-2t \gt $$
Now, $$ x=1+2t \\ \implies dx= 2 dt \\ y=1+4t \\ \implies dy=4 dt \\ z=2-2t \\ \implies dz=-2dt $$
Substitute all the above values in the given integral as follows:
$$\
\int_{0}^1(1+2t) \space (2 -2t)2 dt+(1+2t) \space (2-2t)4d+(1+2t)(1+4t)-2dt=-2$$
