## Thomas' Calculus 13th Edition

$$49$$
We know that the line equation can be defined as: $$r(t)=r_0+kt =(0,0,0)+t \lt 2, 3, -6 \gt=\lt 2t, 3t, -6t \gt$$ So, $$x=2t \\ dx= 2 dt \\y=3t \\ dy=3 dt$$ and $$z=-6t \implies dz=-6 \space dt$$ Substitute all the above values in the given integral as follows: $$\int_{0}^14t(2 dt)+6t (3 dt)-12 t(-6 dt)=49$$