Answer
$3 \sqrt{14}$
Work Step by Step
Since, we have $ds=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
or, $ds=\sqrt{(-1)^2+( -3 )^2+(-2)^2} dt \implies ds= \sqrt {14} dt$
Now, the line integral is: $\int_C (x+y+z) ds=\int_{0}^{1} (1-t+2-3t+3-2t) (\sqrt{14}) dt$
or, $=\sqrt{14}\int_{0}^{1} (2-2t) dt$
or, $=3 \sqrt {14} (2-1)$
or, $=3 \sqrt{14}$
