1. **State the problem:** We need to find the work done by the force field $\vec{F} = 3xy \hat{i} - 5z \hat{j} + 10x \hat{k}$ on a particle moving along the curve defined by $x = t + 1$, $y = 2t$, $z = t$ from $t=1$ to $t=2$. 2. **Formula for work done by a force along a curve:** $$ W = \int_C \vec{F} \cdot d\vec{r} $$ where $d\vec{r} = \frac{d\vec{r}}{dt} dt$ is the differential displacement vector along the curve. 3. **Parameterize the force field along the curve:** Substitute $x = t+1$, $y = 2t$, $z = t$ into $\vec{F}$: $$ \vec{F}(t) = 3(t+1)(2t) \hat{i} - 5(t) \hat{j} + 10(t+1) \hat{k} = 6t(t+1) \hat{i} - 5t \hat{j} + 10(t+1) \hat{k} $$ 4. **Find $d\vec{r}/dt$:** $$ \frac{d\vec{r}}{dt} = \frac{d}{dt}[(t+1) \hat{i} + 2t \hat{j} + t \hat{k}] = 1 \hat{i} + 2 \hat{j} + 1 \hat{k} $$ 5. **Compute the dot product $\vec{F}(t) \cdot \frac{d\vec{r}}{dt}$:** $$ 6t(t+1)(1) + (-5t)(2) + 10(t+1)(1) = 6t^2 + 6t - 10t + 10t + 10 = 6t^2 + 6t + 10 $$ 6. **Set up the integral for work:** $$ W = \int_1^2 (6t^2 + 6t + 10) dt $$ 7. **Integrate:** $$ \int (6t^2 + 6t + 10) dt = 2t^3 + 3t^2 + 10t + C $$ 8. **Evaluate from $t=1$ to $t=2$:** $$ W = [2(2)^3 + 3(2)^2 + 10(2)] - [2(1)^3 + 3(1)^2 + 10(1)] $$ $$ = [2 \times 8 + 3 \times 4 + 20] - [2 + 3 + 10] = (16 + 12 + 20) - 15 = 48 - 15 = 33 $$ **Final answer:** The work done is $33$ units.