Given $\overline{\mathrm{a}} \cdot \overline{\mathrm{r}}=3$
$\overline{\mathrm{a}} \times \overline{\mathrm{r}}=\overline{\mathrm{b}}$
Let $\overline{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}$
$\begin{aligned}
\overline{\mathrm{a}} \times \overline{\mathrm{r}} & =\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
1 & 1 & 1 \\
x & y & \mathrm{z}
\end{array}\right| \\
& =(\mathrm{z}-y) \hat{\mathrm{i}}-\hat{\mathrm{j}}(z-x)+\hat{\mathrm{k}}(y-x)
\end{aligned}$
Given $\overline{\mathrm{a}} \times \overline{\mathrm{r}}=\overline{\mathrm{b}}$
$\therefore \quad(z-y) \hat{\mathrm{i}}-(\mathrm{z}-x) \hat{\mathrm{j}}+(y-x) \hat{\mathrm{k}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$
Comparing
$\begin{aligned}
& \mathrm{z}-y=0 ... (i)\\
& \mathrm{z}-x=-1 ... (ii)\\
& y-x=-1 ... (iii)
\end{aligned}$
Also, $\overline{\mathrm{a}} \cdot \overline{\mathrm{r}}=3$
$\begin{aligned}
& (\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})(x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}})=3 \\
& x+y+z=3
\end{aligned}$
Solving equations (i), (ii), (iii) and (iv), we get
$\begin{aligned}
& x=\frac{5}{3}, y=\frac{2}{3}, \mathrm{z}=\frac{2}{3} \\
\therefore \quad \overline{\mathrm{r}} & =\frac{5}{3} \hat{\mathrm{i}}+\frac{2}{3} \hat{\mathrm{j}}+\frac{2}{3} \hat{\mathrm{k}}
\end{aligned}$