Let $\overline{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}$ then
$$
\begin{aligned}
& \overline{\mathrm{r}} \times \overline{\mathrm{a}}=\overline{\mathrm{b}} \times \overline{\mathrm{a}} \quad \Rightarrow(\overline{\mathrm{r}}-\overline{\mathrm{b}}) \times \overline{\mathrm{a}}=\overline{0} \\
& \therefore \quad\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
x & y-2 & \mathrm{z}+1 \\
1 & 1 & 0
\end{array}\right|=\overline{0} \\
& \Rightarrow(-\mathrm{z}-1) \hat{\mathrm{i}}-(-\mathrm{z}-1) \hat{\mathrm{j}}+(x-y+2) \hat{\mathrm{k}}=\overline{0} \\
& \Rightarrow \mathrm{z}=-1, x-y=-2
\end{aligned}
$$

Now, $\overline{\mathrm{r}} \times \overline{\mathrm{b}}=\overline{\mathrm{a}} \times \overline{\mathrm{b}}=(\overline{\mathrm{r}}-\overline{\mathrm{a}}) \times \overline{\mathrm{b}}=\overline{0}$
$$
\begin{aligned}
\therefore \quad & \left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
x-1 & y-1 & z \\
0 & 2 & -1
\end{array}\right|=\overline{0} \\
& \Rightarrow(1-y-2 \mathrm{z}) \hat{\mathrm{i}}-(1-x) \hat{\mathrm{j}}+(2 x-2) \hat{\mathrm{k}}=\overline{0} \\
& \Rightarrow 1-y-2 z=0, x=1
\end{aligned}
$$

Solving (i) and (ii), we get
$$
\begin{aligned}
& x=1, y=3, z=-1 \\
\therefore \quad & \bar{r}=\hat{i}+3 \hat{j}-\hat{k} \\
& |\bar{r}|=\sqrt{1+9+1}=\sqrt{11} \\
\therefore \quad & \frac{\bar{r}}{\left|-\frac{\mathrm{r}}{-}\right|}=\frac{\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{\sqrt{11}}
\end{aligned}
$$