Answer
$49$
Work Step by Step
We know that the line equation is defined as: $r(t)=r_0+kt$
Thus, $r_0+kt=(0,0,0)+t \lt 2, 3, -6 \gt=\lt 2t, 3t, -6t \gt$
So, $x=2t \implies dx= 2 dt$; $y=3t \implies dy=3 dt$ and $z=-6t \implies dz=-6dt$
We need to substitute all the values in the given integral.
$\int_{0}^14t(2 dt)+6t (3 dt)-12 t(-6 dt)=49$
