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