The parametric equation vec(c)(t)=(6t,-t+8,3t+4) will parameterize the line segment c from (0,8,4) to (6,7,7) as t increases from t=0 to t=1.
If vec(F)(x,y,z) is the vector field you want to integrate over this line segment, the way to calculate the line integral is to calculate int_{0}^{1} vec(F)(vec(c)(t))\cdot vec(c)'(t)\ dt, where vec(F)(vec(c)(t))\cdot vec(c)'(t) is the dot product of vec(F)(vec(c)(t)) with vec(c)'(t).
For example, if vec(F)(x,y,z)=(x+y,y^2+z,xyz), then vec(F)(vec(c)(t))=vec(F)(6t,-t+8,3t+4)
=(5t+8,(-t+8)^2+3t+4,6t(-t+8)(3t+4))
=(5t+8,t^2-13t+68,-18t^3+120t^2+192t).
Hence,
vec(F)(vec(c)(t))\cdot vec(c)'(t)
=(5t+8,t^2-13t+68,-18t^3+120t^2+192t)\cdot (6,-1,3)
=30t+48-t^2+13t-68-54t^3+360t^2+576t
=-54t^3+359t^2+619t-20,
and
int_{0}^{1} vec(F)(vec(c)(t))\cdot vec(c)'(t)\ dt=int_{0}^[1}(-54t^3+359t^2+619t-20)\ dt
=-27/2 t^4+359/3 t^3+619/2 t^2-20t|_{t=0}^{t=1}
=-27/2+359/3+619/2-20=1187/3=395.666666...
You might wonder whether you would get the same answer if you used a different parameterization, such as vec(c)(t)=(3t,-1/2 t+8,3/2 t+4) for 0\leq t\leq 2. The answer is "yes", and you can do the calculation to check it.
If you want to prove this fact in the general case, you have to think about what happens when you "change variables" in the general case. The formula for changing variables in a general case can be written as int_{a}^{b} f(g(x))g'(x)\ dx=int_{g(a)}^{g(b)} f(u)\ du, where u=g(x) is the substitution, du=g'(x)\ dx, u=g(a) when x=a, and u=g(b) when x=b.
