A constant force $\overrightarrow{\mathrm{F}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ newton has a displacement $\vec{r}=2 \hat{i}-3 \hat{j}-3 \hat{k}$ metre in 2 second. The work done and the power are respectively

The work done by a constant force is computed using the dot product of force and displacement vectors. Given $\overrightarrow{F} = 3 \hat{i} - 2 \hat{j} - \hat{k}\, \text{N}$ and $\overrightarrow{r} = 2 \hat{i} - 3 \hat{j} - 3 \hat{k}\, \text{m}$, the work is $\overrightarrow{F} \cdot \overrightarrow{r} = (3)(2) + (-2)(-3) + (-1)(-3) = 6 + 6 + 3 = 15\, \text{J}$.

Power is the rate of doing work, calculated as $P = \frac{W}{t}$. With $W = 15\, \text{J}$ and $t = 2\, \text{s}$, power is $\frac{15}{2} = 7.5\, \text{W}$.

Option B corresponds to these values.