Position vector of starting point $(2,10,1)$ of particle is given as
$$
\mathbf{r}_{1}=2 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}+\hat{\mathbf{k}}
$$
If final co-ordinate of the particle be $(x, y, z)$, then its position vector
$$
\begin{gathered}
\mathbf{r}_{2}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\
\therefore \text { Displacement, } \Delta \mathbf{r}=\mathbf{r}_{2}-\mathbf{r}_{1} \\
8 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}-(2 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \\
\Rightarrow \quad 8 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}=(x-2) \hat{\mathbf{i}}+(y-10) \hat{\mathbf{j}}+(z-1) \hat{\mathbf{k}}
\end{gathered}
$$
Comparing the coefficients of $\hat{\mathbf{i}}, \hat{\mathfrak{j}}$ and $\hat{\mathbf{k}}$, we get
$$
\begin{aligned}
& x-2=8, y-10=-2, z-1=1 \\
\Rightarrow & x=10, y=8, z=2
\end{aligned}
$$
$\therefore$ Final co-ordinate $=(10,8,2)$