Answer
$2$
Work Step by Step
$I=\int_{C_1}(y+z) dx+(x+z) dy+(x+y) dz+\int_{C_2}(y+z) dx+(x+z) dy+(x+y) dz$
The parametric representation of line $C_1$ is:$x=t; y=0; z=t$ and the parametric representation of line $C_2$ is:$x=-t+1; y=t; z=1+t$
Thus, we have $I=\int_{0}^{1} (0+t) dt +(x+z)(0) +(t+0)dt+\int_{0}^{1} [t+(1+t)](-dt) +[(-t+1)+(1+t)] dt +[(-t+1)+t]dt$
$I=\int_0^1 2t dt+\int_{0}^{1} (-2t-1) dt+2 dt+1 dt=\int_0^1 2t dt+\int_{0}^{1} -2t+2 dt=[t^2]_0^1+[-t^2+2t]_{0}^{1}$
Hence, we have $I=(1)^2-0+1=2$
