Answer
$\dfrac{35}{3}$
Work Step by Step
$I=\int_{C}z^2 dx+x^2 dy+y^2 dz$
$=\int_{0}^{1} (2t)^2(3 dt) +(1+3t)^2 (1 dt) +(t)^2 (2 dt)$
$=\int_{0}^{1} (12t^2) dt +(1+6t+9t^2) dt +(2t)^2 dt$
$=\int_{0}^{1} 23t^2+6t+1 dt$
$=[\dfrac{23t^3}{3}+3t^2+t]_0^1$
$=\dfrac{35}{3}$