The work done $W = \int\limits_{c} \bar{F} d\bar{r}$

Now

$\bar{F}d\bar{r} = (3xy\hat{i} - 5z\hat{j} + 10x\hat{k})(dx\hat{i} + dy\hat{j} + dz\hat{k}) = 3xydx - 5zdy + 10xdz$

Now , $x = t^2 + 1, dx = 2tdt \\ y = 2t^2, dy = 4tdt \\ z = t^3, dz = 3, z = t^3$

$\bar{F}d\bar{r} = 3(t^2 + 1)(2t^2)(2tdt) - 5t^3(4tdt) + 10(t^2 + 1)(3t^2dt) \\ \therefore \bar{F}d\bar{r} = (12t^5 + 12t^3 + 10t^4 + 30t^2)dt $

$\therefore W = \int\bar{F}d\bar{r} = \int\limits_{1}^{2}(12t^5 + 12t^3 + 10t^4 + 30t^2)dt \\ \therefore W = \bigg[12\frac{t^6}{6} + 10\frac{t^5}{5} + 12 \frac{t^4}{4} + 30\frac{t^3}{3}\bigg]_{1}^{2} \\ \therefore W = [2t^6 + 2t^5 + 3t^4 + 10t^3]_{1}^{2} \\ \therefore W = [128 + 64 + 48 + 20 - 2 -2 - 3 - 10] \\ \therefore W = 303$